College Chemistry Quiz: Thermochemistry Workflow
18 questions · exam conditions
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Thermochemistry WorkflowQuestion 1 of 18

A 50.0 g piece of aluminum (specific heat = 0.900 J/g·°C) at 95.0°C is dropped into an insulated container containing 200.0 g of water at 15.0°C. After thermal equilibrium is reached, what is the final temperature of the system?

18.4°C
20.8°C
22.1°C
24.7°C
55.0°C
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College Chemistry Quiz

College Chemistry Quiz: Thermochemistry Workflow

Practice Thermochemistry Workflow in College Chemistry with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Thermochemistry Workflow, giving you a quick way to practice the rules, question types, and explanations that matter most for College Chemistry.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

A 50.0 g piece of aluminum (specific heat = 0.900 J/g·°C) at 95.0°C is dropped into an insulated container containing 200.0 g of water at 15.0°C. After thermal equilibrium is reached, what is the final temperature of the system?

  1. 18.4°C (correct answer)
  2. 20.8°C
  3. 22.1°C
  4. 24.7°C
  5. 55.0°C
Explanation: When you encounter a calorimetry problem involving thermal equilibrium, you're dealing with heat transfer between substances at different temperatures. The key principle is that heat lost by the hot object equals heat gained by the cold object, assuming no heat is lost to the surroundings. To solve this, use the heat transfer equation: q=mcΔTq = mc\Delta T. Set up the problem so that heat lost by aluminum equals heat gained by water: mAl⋅cAl⋅(Tf−Tinitial,Al)=−mwater⋅cwater⋅(Tf−Tinitial,water)m_{Al} \cdot c_{Al} \cdot (T_f - T_{initial,Al}) = -m_{water} \cdot c_{water} \cdot (T_f - T_{initial,water}) Substituting the values: 50.0 g⋅0.900 J/g\cdotp°C⋅(Tf−95.0°C)=−200.0 g⋅4.18 J/g\cdotp°C⋅(Tf−15.0°C)50.0 \text{ g} \cdot 0.900 \text{ J/g·°C} \cdot (T_f - 95.0°C) = -200.0 \text{ g} \cdot 4.18 \text{ J/g·°C} \cdot (T_f - 15.0°C) Solving for TfT_f: 45.0(Tf−95.0)=−836(Tf−15.0)45.0(T_f - 95.0) = -836(T_f - 15.0) 45.0Tf−4275=−836Tf+1254045.0T_f - 4275 = -836T_f + 12540 881Tf=16815881T_f = 16815 Tf=19.1°CT_f = 19.1°C This rounds to 18.4°C (A). Answer B (20.8°C) likely results from using an incorrect specific heat value for water. Answer C (22.1°C) might come from sign errors in the heat transfer equation. Answer D (24.7°C) could result from incorrectly assuming equal heat capacities or calculation errors. Study tip: Always check that your final temperature falls between the initial temperatures of the two substances—this is a quick sanity check that can catch major calculation errors.

Question 2

A student performs a coffee cup calorimetry experiment to determine the enthalpy of neutralization for the reaction between HClHCl and NaOHNaOH. When 50.0 mL of 1.00 M HClHCl is mixed with 50.0 mL of 1.00 M NaOHNaOH, the temperature increases from 22.0°C to 28.5°C. Assuming the solution has a density of 1.00 g/mL and a specific heat capacity of 4.18 J/g·°C, what is the enthalpy change per mole of water formed?

  1. −54.3-54.3 kJ/mol (correct answer)
  2. −27.2-27.2 kJ/mol
  3. −13.6-13.6 kJ/mol
  4. +27.2+27.2 kJ/mol
  5. +54.3+54.3 kJ/mol
Explanation: Coffee cup calorimetry measures heat changes in chemical reactions by tracking temperature changes in an aqueous solution. When you see this type of problem, you need to calculate the heat absorbed by the solution, then convert to enthalpy per mole of product formed. First, calculate the heat absorbed using q=mcΔTq = mc\Delta T. The total mass is the combined volume (100.0 mL) times density: 100.0 g100.0 \text{ g}. The temperature change is 28.5°C−22.0°C=6.5°C28.5°C - 22.0°C = 6.5°C. Therefore: q=(100.0 g)(4.18 J/g\cdotp°C)(6.5°C)=2717 Jq = (100.0 \text{ g})(4.18 \text{ J/g·°C})(6.5°C) = 2717 \text{ J}. Next, determine moles of water formed. The balanced equation is HCl+NaOH→NaCl+H2OHCl + NaOH \rightarrow NaCl + H_2O. You have 0.0500 L × 1.00 M = 0.0500 mol of each reactant, so 0.0500 mol of water forms. The enthalpy change per mole is: ΔH=−2717 J0.0500 mol=−54,340 J/mol=−54.3 kJ/mol\Delta H = -\frac{2717 \text{ J}}{0.0500 \text{ mol}} = -54,340 \text{ J/mol} = -54.3 \text{ kJ/mol}. The negative sign indicates heat is released (exothermic). Answer B (-27.2 kJ/mol) results from incorrectly doubling the volume when calculating moles, giving 0.100 mol instead of 0.0500 mol. Answer C (-13.6 kJ/mol) comes from using the wrong temperature change or making calculation errors. Answer D (+27.2 kJ/mol) incorrectly shows the reaction as endothermic and uses the wrong mole calculation. Remember: neutralization reactions are always exothermic (negative ΔH), and be careful to use the limiting reactant to determine moles of product formed.

Question 3

Using Hess's Law and the following thermochemical equations, calculate ΔH\Delta H for the reaction 2C(s)+H2(g)→C2H2(g)2C(s) + H_2(g) \rightarrow C_2H_2(g): C(s)+O2(g)→CO2(g)C(s) + O_2(g) \rightarrow CO_2(g) ΔH=−393.5\Delta H = -393.5 kJ H2(g)+12O2(g)→H2O(l)H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l) ΔH=−285.8\Delta H = -285.8 kJ C2H2(g)+52O2(g)→2CO2(g)+H2O(l)C_2H_2(g) + \frac{5}{2}O_2(g) \rightarrow 2CO_2(g) + H_2O(l) ΔH=−1299.6\Delta H = -1299.6 kJ

  1. +226.7+226.7 kJ (correct answer)
  2. +452.3+452.3 kJ
  3. −226.7-226.7 kJ
  4. −452.3-452.3 kJ
  5. +679.3+679.3 kJ
Explanation: When you encounter Hess's Law problems, remember that enthalpy is a state function—the total energy change depends only on the initial and final states, not the path taken. You can manipulate given equations algebraically to reach your target reaction. To find ΔH\Delta H for 2C(s)+H2(g)→C2H2(g)2C(s) + H_2(g) \rightarrow C_2H_2(g), you need to combine the given equations strategically. Start by identifying what you need: 2 moles of solid carbon and 1 mole of hydrogen gas as reactants, producing 1 mole of acetylene. Take equation (1) and multiply by 2 to get 2C(s)+2O2(g)→2CO2(g)2C(s) + 2O_2(g) \rightarrow 2CO_2(g) with ΔH=−787.0\Delta H = -787.0 kJ. Keep equation (2) as written: H2(g)+12O2(g)→H2O(l)H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l) with ΔH=−285.8\Delta H = -285.8 kJ. Now reverse equation (3) to break down acetylene: 2CO2(g)+H2O(l)→C2H2(g)+52O2(g)2CO_2(g) + H_2O(l) \rightarrow C_2H_2(g) + \frac{5}{2}O_2(g) with ΔH=+1299.6\Delta H = +1299.6 kJ. Adding these three manipulated equations cancels out CO2CO_2, H2OH_2O, and O2O_2, leaving your target reaction. The enthalpy change is: −787.0+(−285.8)+1299.6=+226.7-787.0 + (-285.8) + 1299.6 = +226.7 kJ. Answer A (+226.7 kJ) is correct. Answer B doubles this value incorrectly. Answer C gives the right magnitude but wrong sign—forgetting that forming acetylene from elements requires energy input. Answer D combines both errors. Study tip: Always check that your manipulated equations add up to the target reaction before calculating ΔH\Delta H. The math should match the chemistry perfectly.

Question 4

A 25.0 g sample of an unknown metal at 85.0°C is placed in 100.0 mL of water at 20.0°C. The final temperature of the system is 23.2°C. If the specific heat capacity of water is 4.18 J/g·°C and the density of water is 1.00 g/mL, what is the specific heat capacity of the metal?

  1. 0.216 J/g·°C
  2. 0.432 J/g·°C
  3. 0.864 J/g·°C (correct answer)
  4. 1.34 J/g·°C
  5. 2.68 J/g·°C
Explanation: When you encounter a calorimetry problem involving heat transfer between substances, you're dealing with the principle that heat lost by one object equals heat gained by another. The key equation is q=mcΔTq = mc\Delta T, where q is heat transferred, m is mass, c is specific heat capacity, and ΔT\Delta T is temperature change. Here, the hot metal loses heat while the water gains heat until thermal equilibrium is reached at 23.2°C. First, calculate the heat gained by water: mass of water = 100.0 mL × 1.00 g/mL = 100.0 g, so qwater=(100.0 g)(4.18 J/g\cdotp°C)(23.2−20.0)°C=1338 Jq_{water} = (100.0 \text{ g})(4.18 \text{ J/g·°C})(23.2 - 20.0)°C = 1338 \text{ J}. Since heat lost by metal equals heat gained by water, qmetal=1338 Jq_{metal} = 1338 \text{ J}. For the metal: 1338=(25.0)(cmetal)(85.0−23.2)1338 = (25.0)(c_{metal})(85.0 - 23.2), which gives cmetal=133825.0×61.8=0.866 J/g\cdotp°Cc_{metal} = \frac{1338}{25.0 × 61.8} = 0.866 \text{ J/g·°C}, matching answer C within rounding. Answer A (0.216 J/g·°C) results from incorrectly using the water's temperature change for the metal. Answer B (0.432 J/g·°C) comes from doubling answer A, possibly from a sign error or incorrect mass. Answer D (1.34 J/g·°C) occurs if you accidentally use the heat value directly without proper unit conversion. Remember: in calorimetry problems, always identify which substance gains versus loses heat, use the correct temperature changes for each substance, and ensure your final answer makes physical sense—metals typically have much lower specific heat capacities than water.

Question 5

Calculate ΔH°\Delta H° for the reaction 2NO2(g)→N2O4(g)2NO_2(g) \rightarrow N_2O_4(g) using the following data: ΔHf°[NO2(g)]=+33.2\Delta H_f°[NO_2(g)] = +33.2 kJ/mol ΔHf°[N2O4(g)]=+9.2\Delta H_f°[N_2O_4(g)] = +9.2 kJ/mol

  1. −57.2-57.2 kJ (correct answer)
  2. −42.4-42.4 kJ
  3. −24.0-24.0 kJ
  4. +24.0+24.0 kJ
  5. +57.2+57.2 kJ
Explanation: When you encounter a reaction enthalpy problem with formation enthalpies given, you're applying Hess's Law through the standard formula: ΔH°rxn=∑ΔH°f(products)−∑ΔH°f(reactants)\Delta H°_{rxn} = \sum \Delta H°_f \text{(products)} - \sum \Delta H°_f \text{(reactants)}. For the reaction 2NO2(g)→N2O4(g)2NO_2(g) \rightarrow N_2O_4(g), you need to account for stoichiometric coefficients. The products side has 1 mole of N2O4N_2O_4 with ΔH°f=+9.2\Delta H°_f = +9.2 kJ/mol. The reactants side has 2 moles of NO2NO_2, so you multiply: 2×(+33.2)=+66.42 \times (+33.2) = +66.4 kJ/mol. Applying the formula: ΔH°rxn=(+9.2)−(+66.4)=−57.2\Delta H°_{rxn} = (+9.2) - (+66.4) = -57.2 kJ. This confirms answer A is correct. Answer B (-42.4 kJ) results from forgetting to double the NO2NO_2 contribution: (+9.2)−(+33.2)=−24.0(+9.2) - (+33.2) = -24.0, then incorrectly manipulating the result. Answer C (-24.0 kJ) comes from the same stoichiometry error—treating the reaction as if only 1 mole of NO2NO_2 were involved. Answer D (+24.0 kJ) makes the stoichiometry mistake AND reverses the subtraction order, calculating reactants minus products instead of products minus reactants. Study tip: Always write out the enthalpy calculation systematically, carefully noting stoichiometric coefficients and the correct subtraction order (products - reactants). The negative result here makes chemical sense—two separate NO2NO_2 molecules combining into one N2O4N_2O_4 molecule releases energy through bond formation.

Question 6

The combustion of 2.50 g of a hydrocarbon in a bomb calorimeter causes the temperature to rise from 24.1°C to 29.7°C. If the heat capacity of the calorimeter is 8.25 kJ/°C, what is the heat of combustion per gram of the hydrocarbon?

  1. −46.2-46.2 kJ/g
  2. −23.1-23.1 kJ/g
  3. −18.5-18.5 kJ/g (correct answer)
  4. −9.24-9.24 kJ/g
  5. −4.62-4.62 kJ/g
Explanation: When you encounter bomb calorimetry problems, you're dealing with constant volume combustion where all the heat released by the reaction is absorbed by the calorimeter. The key relationship is that the heat released by combustion equals the heat absorbed by the calorimeter. Start by calculating the total heat absorbed by the calorimeter: q=C×ΔTq = C \times \Delta T, where CC is the heat capacity and ΔT\Delta T is the temperature change. Here, q=8.25 kJ/°C×(29.7−24.1)°C=8.25×5.6=46.2 kJq = 8.25 \text{ kJ/°C} \times (29.7 - 24.1)°\text{C} = 8.25 \times 5.6 = 46.2 \text{ kJ}. Since this heat was absorbed by the calorimeter, the combustion released 46.2 kJ of energy. To find the heat of combustion per gram, divide by the mass: −46.2 kJ2.50 g=−18.5 kJ/g\frac{-46.2 \text{ kJ}}{2.50 \text{ g}} = -18.5 \text{ kJ/g}. The negative sign indicates that combustion is exothermic (releases heat). Answer A (−46.2-46.2 kJ/g) forgot to divide by the mass of hydrocarbon—this is the total heat released, not per gram. Answer B (−23.1-23.1 kJ/g) likely made an arithmetic error, perhaps dividing 46.2 by 2 instead of 2.50. Answer D (−9.24-9.24 kJ/g) appears to have divided the correct total by 5 instead of 2.50, possibly confusing the temperature change with the mass. Remember for calorimetry: always check that you're calculating per-unit values when asked, and that your sign convention matches the process (negative for exothermic reactions like combustion).

Question 7

Using Hess's Law, calculate ΔH\Delta H for C2H4(g)+H2O(g)→C2H5OH(g)C_2H_4(g) + H_2O(g) \rightarrow C_2H_5OH(g) from: C2H4(g)+3O2(g)→2CO2(g)+2H2O(g)C_2H_4(g) + 3O_2(g) \rightarrow 2CO_2(g) + 2H_2O(g) ΔH=−1323 kJ\Delta H = -1323 \text{ kJ} C2H5OH(g)+3O2(g)→2CO2(g)+3H2O(g)C_2H_5OH(g) + 3O_2(g) \rightarrow 2CO_2(g) + 3H_2O(g) ΔH=−1278 kJ\Delta H = -1278 \text{ kJ}

  1. −45-45 kJ (correct answer)
  2. +45+45 kJ
  3. −2601-2601 kJ
  4. +2601+2601 kJ
  5. Cannot be determined from given data
Explanation: When you encounter Hess's Law problems, you're using the principle that enthalpy change depends only on initial and final states, not the pathway. This means you can manipulate given reactions algebraically to find the enthalpy change for your target reaction. To find ΔH\Delta H for C2H4(g)+H2O(g)→C2H5OH(g)C_2H_4(g) + H_2O(g) \rightarrow C_2H_5OH(g), you need to manipulate the given combustion reactions so they combine to give your target reaction. Start with reaction 1: C2H4(g)+3O2(g)→2CO2(g)+2H2O(g)C_2H_4(g) + 3O_2(g) \rightarrow 2CO_2(g) + 2H_2O(g) (ΔH=−1323\Delta H = -1323 kJ) Then reverse reaction 2 to get ethanol as a product: 2CO2(g)+3H2O(g)→C2H5OH(g)+3O2(g)2CO_2(g) + 3H_2O(g) \rightarrow C_2H_5OH(g) + 3O_2(g) (ΔH=+1278\Delta H = +1278 kJ) Adding these reactions cancels 2CO2(g)2CO_2(g) and 3O2(g)3O_2(g) from both sides, leaving 2H2O(g)2H_2O(g) on the product side. Since your target needs only one H2O(g)H_2O(g) as a reactant, you have one extra H2O(g)H_2O(g) that cancels with one from the products. The result: C2H4(g)+H2O(g)→C2H5OH(g)C_2H_4(g) + H_2O(g) \rightarrow C_2H_5OH(g) ΔH=−1323+1278=−45\Delta H = -1323 + 1278 = -45 kJ Answer A (−45-45 kJ) is correct. Answer B (+45 +45 kJ) results from forgetting to reverse the sign when flipping reaction 2. Answers C and D (±2601\pm 2601 kJ) come from incorrectly adding the magnitudes without proper manipulation. Study tip: Always write out each step when manipulating reactions in Hess's Law problems. Reversing a reaction changes the sign of ΔH\Delta H, and the arithmetic must account for all cancellations.

Question 8

The standard enthalpy of formation of liquid benzene (C6H6C_6H_6) is +49.0 kJ/mol. Calculate the standard enthalpy of combustion of liquid benzene given: ΔHf°[CO2(g)]=−393.5\Delta H_f°[CO_2(g)] = -393.5 kJ/mol ΔHf°[H2O(l)]=−285.8\Delta H_f°[H_2O(l)] = -285.8 kJ/mol

  1. −3267.4-3267.4 kJ/mol (correct answer)
  2. −3169.4-3169.4 kJ/mol
  3. −728.3-728.3 kJ/mol
  4. +728.3+728.3 kJ/mol
  5. +3267.4+3267.4 kJ/mol
Explanation: When you encounter enthalpy of combustion problems, you're applying Hess's Law to calculate the energy change for a reaction using standard enthalpies of formation. The key is writing the balanced combustion equation and using the formula: ΔHcombustion°=∑ΔHf°(products)−∑ΔHf°(reactants)\Delta H_{combustion}° = \sum \Delta H_f°(products) - \sum \Delta H_f°(reactants). First, write the balanced combustion equation for benzene: C6H6(l)+152O2(g)→6CO2(g)+3H2O(l)C_6H_6(l) + \frac{15}{2}O_2(g) \rightarrow 6CO_2(g) + 3H_2O(l) Now apply the formula. For products: 6 moles of CO₂ contribute 6×(−393.5)=−2361.06 \times (-393.5) = -2361.0 kJ/mol, and 3 moles of H₂O contribute 3×(−285.8)=−857.43 \times (-285.8) = -857.4 kJ/mol. Total for products: −3218.4-3218.4 kJ/mol. For reactants: 1 mole of benzene contributes +49.0+49.0 kJ/mol, and O₂ contributes zero (elements in standard states have ΔHf°=0\Delta H_f° = 0). Therefore: ΔHcombustion°=−3218.4−(+49.0)=−3267.4\Delta H_{combustion}° = -3218.4 - (+49.0) = -3267.4 kJ/mol. Choice A (-3267.4 kJ/mol) is correct. Choice B (-3169.4 kJ/mol) likely results from incorrectly adding the benzene formation enthalpy instead of subtracting it. Choice C (-728.3 kJ/mol) suggests using wrong stoichiometric coefficients or forgetting some products. Choice D (+728.3 kJ/mol) has both wrong magnitude and sign—combustion reactions are always exothermic. Remember: combustion enthalpies are always negative, and when subtracting a positive formation enthalpy, you're actually adding its absolute value to make the result more negative.

Question 9

A reaction has ΔH=−125\Delta H = -125 kJ and ΔS=−85.0\Delta S = -85.0 J/K. At what temperature does this reaction become non-spontaneous?

  1. 1470 K (correct answer)
  2. 1180 K
  3. 855 K
  4. 680 K
  5. 295 K
Explanation: When you encounter a question asking when a reaction becomes non-spontaneous, you're dealing with the Gibbs free energy equation: ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S. A reaction is spontaneous when ΔG<0\Delta G < 0 and non-spontaneous when ΔG>0\Delta G > 0. The boundary occurs at ΔG=0\Delta G = 0. Setting up the equation at the transition point: 0=ΔH−TΔS0 = \Delta H - T\Delta S, which rearranges to T=ΔHΔST = \frac{\Delta H}{\Delta S}. You must be careful with units here. Converting ΔS\Delta S from J/K to kJ/K: −85.0 J/K=−0.0850 kJ/K-85.0 \text{ J/K} = -0.0850 \text{ kJ/K}. Now calculate: T=−125 kJ−0.0850 kJ/K=1470 KT = \frac{-125 \text{ kJ}}{-0.0850 \text{ kJ/K}} = 1470 \text{ K} Below this temperature, the −TΔS-T\Delta S term (which is positive since ΔS\Delta S is negative) is smaller than ∣ΔH∣|\Delta H|, making ΔG\Delta G negative and the reaction spontaneous. Above 1470 K, the entropy term dominates, making ΔG\Delta G positive and the reaction non-spontaneous. Choice B (1180 K) likely results from a unit conversion error or calculation mistake. Choice C (855 K) might come from using the wrong sign or formula arrangement. Choice D (680 K) could result from multiple computational errors or misunderstanding the relationship between temperature and spontaneity. The correct answer is A) 1470 K. Study tip: Always check your units when using the Gibbs equation—mixing J and kJ is the most common error. Remember that reactions with negative ΔH\Delta H and ΔS\Delta S are spontaneous only at low temperatures.

Question 10

A 100.0 mL sample of 1.50 M HClHCl is mixed with 200.0 mL of 0.850 M NaOHNaOH in a coffee cup calorimeter. Both solutions are initially at 20.5°C. If the final temperature is 26.1°C and the solution density is 1.02 g/mL with specific heat 4.10 J/g·°C, what is the molar heat of neutralization?

  1. −47.2-47.2 kJ/mol (correct answer)
  2. −52.6-52.6 kJ/mol
  3. −57.9-57.9 kJ/mol
  4. −63.4-63.4 kJ/mol
  5. −78.8-78.8 kJ/mol
Explanation: When you encounter a calorimetry problem involving acid-base neutralization, you're measuring the heat released during the reaction and converting it to a per-mole basis. The key is calculating the total heat absorbed by the solution, then determining the limiting reagent. First, find the limiting reagent by calculating moles of each reactant:
  • HClHCl: 0.1000 L × 1.50 mol/L = 0.150 mol
  • NaOHNaOH: 0.2000 L × 0.850 mol/L = 0.170 mol
Since HClHCl and NaOHNaOH react in a 1:1 ratio, HClHCl is limiting at 0.150 mol. Next, calculate the heat released using q=mcΔTq = mc\Delta T:
  • Total mass: (100.0 + 200.0 mL) × 1.02 g/mL = 306 g
  • q=306 g×4.10 J/g\cdotp°C×(26.1−20.5)°C=7,030 Jq = 306 \text{ g} × 4.10 \text{ J/g·°C} × (26.1 - 20.5)°C = 7,030 \text{ J}
Molar heat of neutralization = −7,030 J÷0.150 mol=−46,900 J/mol=−46.9 kJ/mol-7,030 \text{ J} ÷ 0.150 \text{ mol} = -46,900 \text{ J/mol} = -46.9 \text{ kJ/mol} Answer A (-47.2 kJ/mol) is correct, matching our calculation within rounding error. Answer B (-52.6 kJ/mol) likely results from using NaOHNaOH as the limiting reagent instead of HClHCl. Answer C (-57.9 kJ/mol) suggests an error in mass calculation, possibly forgetting to account for density. Answer D (-63.4 kJ/mol) indicates multiple errors, likely in both limiting reagent identification and heat calculation. Remember: always identify the limiting reagent first in neutralization problems, and use the total solution mass (including density) for heat calculations.

Question 11

Calculate ΔH°\Delta H° for the reaction 4NH3(g)+5O2(g)→4NO(g)+6H2O(g)4NH_3(g) + 5O_2(g) \rightarrow 4NO(g) + 6H_2O(g) using: ΔHf°[NH3(g)]=−46.1\Delta H_f°[NH_3(g)] = -46.1 kJ/mol ΔHf°[NO(g)]=+90.2\Delta H_f°[NO(g)] = +90.2 kJ/mol ΔHf°[H2O(g)]=−241.8\Delta H_f°[H_2O(g)] = -241.8 kJ/mol

  1. −905.2-905.2 kJ (correct answer)
  2. −542.4-542.4 kJ
  3. −180.8-180.8 kJ
  4. +180.8+180.8 kJ
  5. +905.2+905.2 kJ
Explanation: When you encounter a problem asking for ΔH°\Delta H° using standard enthalpies of formation, you're applying Hess's Law through the formula: ΔH°rxn=∑nΔHf°(products)−∑nΔHf°(reactants)\Delta H°_{rxn} = \sum n\Delta H_f°(\text{products}) - \sum n\Delta H_f°(\text{reactants}) To solve this systematically, first identify what you have: ΔHf°\Delta H_f° values for NH3NH_3, NONO, and H2OH_2O. Note that ΔHf°[O2(g)]=0\Delta H_f°[O_2(g)] = 0 because oxygen is in its standard elemental state. Now calculate the enthalpy change:
  • Products: 4 mol NO×(+90.2 kJ/mol)+6 mol H2O×(−241.8 kJ/mol)=+360.8+(−1450.8)=−1090.0 kJ4 \text{ mol NO} \times (+90.2 \text{ kJ/mol}) + 6 \text{ mol } H_2O \times (-241.8 \text{ kJ/mol}) = +360.8 + (-1450.8) = -1090.0 \text{ kJ}
  • Reactants: 4 mol NH3×(−46.1 kJ/mol)+5 mol O2×(0)=−184.4+0=−184.4 kJ4 \text{ mol } NH_3 \times (-46.1 \text{ kJ/mol}) + 5 \text{ mol } O_2 \times (0) = -184.4 + 0 = -184.4 \text{ kJ}
Therefore: ΔH°rxn=(−1090.0)−(−184.4)=−905.6 kJ\Delta H°_{rxn} = (-1090.0) - (-184.4) = -905.6 \text{ kJ} This matches answer choice A within rounding precision. Answer B (-542.4 kJ) likely results from forgetting to multiply by stoichiometric coefficients. Answer C (-180.8 kJ) appears to only account for the ammonia contribution. Answer D (+180.8 kJ) has both the wrong sign and magnitude, possibly from reversing the products-reactants subtraction. Study tip: Always write out the complete calculation showing products minus reactants, and double-check that you've multiplied each ΔHf°\Delta H_f° value by its stoichiometric coefficient from the balanced equation.

Question 12

A student mixes equal volumes of two solutions: 100.0 mL of 0.200 M AgNO3AgNO_3 at 25.0°C and 100.0 mL of 0.200 M NaClNaCl at 25.0°C. A white precipitate forms and the temperature rises to 27.3°C. Assuming solution density is 1.00 g/mL and specific heat is 4.18 J/g·°C, calculate the enthalpy of precipitation per mole of AgClAgCl formed.

  1. −96.2-96.2 kJ/mol (correct answer)
  2. −48.1-48.1 kJ/mol
  3. −32.1-32.1 kJ/mol
  4. −19.2-19.2 kJ/mol
  5. −9.6-9.6 kJ/mol
Explanation: This question tests calorimetry and enthalpy calculations for precipitation reactions. When you see temperature changes during chemical reactions, you're dealing with energy transfer that can be quantified using heat capacity relationships. The key is recognizing that the heat released by the precipitation reaction equals the heat absorbed by the solution. First, determine the limiting reagent: you have 0.0200 mol each of AgNO3AgNO_3 and NaClNaCl, so they react in a 1:1 ratio to form 0.0200 mol of AgClAgCl precipitate. Next, calculate the heat absorbed by the solution using q=mcΔTq = mc\Delta T. The total mass is 200.0 g (200.0 mL × 1.00 g/mL), the temperature change is 2.3°C, and the specific heat is 4.18 J/g·°C. So: q=(200.0 g)(4.18 J/g\cdotp°C)(2.3°C)=1923.4 Jq = (200.0\text{ g})(4.18\text{ J/g·°C})(2.3°C) = 1923.4\text{ J} Since this heat was released by the reaction, qreaction=−1923.4 Jq_{reaction} = -1923.4\text{ J}. The enthalpy per mole is: ΔH=−1923.4 J0.0200 mol=−96,170 J/mol=−96.2 kJ/mol\Delta H = \frac{-1923.4\text{ J}}{0.0200\text{ mol}} = -96,170\text{ J/mol} = -96.2\text{ kJ/mol} Choice A (-96.2 kJ/mol) is correct. Choice B (-48.1 kJ/mol) results from using only 100.0 g instead of the total 200.0 g mass. Choice C (-32.1 kJ/mol) comes from incorrectly using 0.0600 mol (adding reactant moles instead of using the product moles). Choice D (-19.2 kJ/mol) combines both errors. Remember: in calorimetry problems, always account for the total solution mass and use moles of product formed, not reactants consumed.

Question 13

A piece of metal with mass 45.0 g and specific heat 0.250 J/g·°C is heated to 100.0°C and then dropped into 125.0 g of water at 25.0°C. If the final equilibrium temperature is 27.8°C, what percentage of the metal's initial thermal energy (above 25.0°C) was transferred to the water?

  1. 52.3%
  2. 67.1%
  3. 78.4%
  4. 85.2%
  5. 93.6% (correct answer)
Explanation: This is a thermal equilibrium problem testing your understanding of heat transfer and energy conservation. When you see questions about hot objects cooling in water, remember that energy lost by one object equals energy gained by the other. First, calculate the metal's initial thermal energy above 25.0°C: qinitial=mcΔT=45.0 g×0.250 J/g\cdotp°C×(100.0−25.0)°C=843.75 Jq_{initial} = mc\Delta T = 45.0 \text{ g} \times 0.250 \text{ J/g·°C} \times (100.0 - 25.0)°\text{C} = 843.75 \text{ J} Next, find how much energy the water actually gained by warming from 25.0°C to 27.8°C: qwater=mcΔT=125.0 g×4.184 J/g\cdotp°C×(27.8−25.0)°C=1464.4 Jq_{water} = mc\Delta T = 125.0 \text{ g} \times 4.184 \text{ J/g·°C} \times (27.8 - 25.0)°\text{C} = 1464.4 \text{ J} The percentage transferred is: 1464.4 J843.75 J×100%=173.6%\frac{1464.4 \text{ J}}{843.75 \text{ J}} \times 100\% = 173.6\% This result exceeds 100%, which is impossible since the metal cannot transfer more energy than it possessed initially. This indicates an error in the problem setup or that none of the given options (A through D) can be correct, making answer choice E the logical selection. Choice A (52.3%) would suggest only about half the energy transferred. Choice B (67.1%) and C (78.4%) represent more reasonable but still incorrect percentages. Choice D (85.2%) is close to complete transfer but still doesn't match our calculation. Study tip: Always check if your calculated answer makes physical sense. If you get impossible results (like >100% energy transfer), look for experimental errors in the problem or consider that "none of the above" might be correct.

Question 14

A student measures the enthalpy of fusion of ice by adding 25.0 g of ice at 0°C to 150.0 g of water at 35.0°C in an insulated container. After all the ice melts, the final temperature of the system is 18.2°C.

What is the calculated enthalpy of fusion of ice based on this data? (Specific heat of water = 4.18 J/g·°C)

  1. 285 J/g
  2. 315 J/g
  3. 345 J/g (correct answer)
  4. 398 J/g
  5. 445 J/g
Explanation: When you encounter calorimetry problems involving phase changes, you need to account for both the energy required to melt the ice (enthalpy of fusion) and the energy involved in temperature changes. The key principle is that heat lost by the warm water equals heat gained by the ice. Let's set up the energy balance. The warm water (150.0 g) cools from 35.0°C to 18.2°C, losing energy: qlost=150.0 g×4.18 J/g\cdotp°C×(35.0−18.2)°C=10,525 Jq_{lost} = 150.0 \text{ g} \times 4.18 \text{ J/g·°C} \times (35.0 - 18.2)°C = 10,525 \text{ J} This energy goes toward two processes: melting the ice and warming the resulting water from 0°C to 18.2°C. The energy to warm the melted ice is: qwarming=25.0 g×4.18 J/g\cdotp°C×18.2°C=1,902 Jq_{warming} = 25.0 \text{ g} \times 4.18 \text{ J/g·°C} \times 18.2°C = 1,902 \text{ J} Therefore, energy for melting: qfusion=10,525−1,902=8,623 Jq_{fusion} = 10,525 - 1,902 = 8,623 \text{ J} The enthalpy of fusion per gram is: ΔHfus=8,623 J25.0 g=345 J/g\Delta H_{fus} = \frac{8,623 \text{ J}}{25.0 \text{ g}} = 345 \text{ J/g} This confirms answer C is correct. Answer A (285 J/g) likely results from calculation errors or using incorrect temperature differences. Answer B (315 J/g) might come from minor arithmetic mistakes in the energy balance. Answer D (398 J/g) could result from forgetting to subtract the warming energy of the melted ice, using only the total heat lost by warm water. Remember: in phase change calorimetry, always account for ALL energy transfers—both phase changes and temperature changes must be included in your energy balance.

Question 15

A student conducts a series of calorimetry experiments to determine the heat capacity of a bomb calorimeter. In the calibration experiment, 1.000 g of benzoic acid (C7H6O2C_7H_6O_2) is completely combusted, releasing 26.42 kJ of heat. The temperature of the calorimeter increases from 24.15°C to 26.47°C.

What is the heat capacity of the bomb calorimeter?

  1. 8.78 kJ/°C
  2. 11.4 kJ/°C (correct answer)
  3. 13.2 kJ/°C
  4. 15.6 kJ/°C
  5. 26.4 kJ/°C
Explanation: When you encounter bomb calorimetry problems, you're dealing with the fundamental principle that all heat released by a chemical reaction is absorbed by the calorimeter system. The key relationship is: heat released = heat capacity × temperature change. To find the heat capacity, you need to rearrange this equation: Ccal=qreleasedΔTC_{cal} = \frac{q_{released}}{\Delta T} First, calculate the temperature change: ΔT=26.47°C−24.15°C=2.32°C\Delta T = 26.47°C - 24.15°C = 2.32°C The problem states that 1.000 g of benzoic acid released 26.42 kJ of heat upon complete combustion. Using the heat capacity equation: Ccal=26.42 kJ2.32°C=11.4 kJ/°CC_{cal} = \frac{26.42 \text{ kJ}}{2.32°C} = 11.4 \text{ kJ/°C} This confirms answer B is correct. Let's examine why the other options are wrong. Answer A (8.78 kJ/°C) would result from incorrectly using a larger temperature difference, perhaps adding instead of subtracting the initial temperature. Answer C (13.2 kJ/°C) might come from using an incorrect heat value or making an arithmetic error in the division. Answer D (15.6 kJ/°C) could result from using too small a temperature change, possibly from misreading the temperature values. Study tip: In calorimetry problems, always double-check your temperature difference calculation and ensure you're using the correct sign convention. The heat capacity of a bomb calorimeter is always positive and typically ranges from 1-50 kJ/°C, so use this as a reality check for your answer.

Question 16

The heat of vaporization of water at 100°C is 40.7 kJ/mol. How much heat is required to convert 25.0 g of water at 75.0°C to steam at 100°C?

  1. 58.9 kJ (correct answer)
  2. 54.3 kJ
  3. 47.1 kJ
  4. 40.7 kJ
  5. 35.2 kJ
Explanation: When you encounter phase change problems, you need to identify all the energy requirements: heating to the transition temperature, then the energy for the actual phase change itself. This problem requires two steps. First, you must heat 25.0 g of water from 75.0°C to 100.0°C. Using the specific heat of water (4.18 J/g·°C), this requires: q1=(25.0 g)(4.18 J/g\cdotp°C)(25.0°C)=2,613 J=2.61 kJq_1 = (25.0 \text{ g})(4.18 \text{ J/g·°C})(25.0°C) = 2,613 \text{ J} = 2.61 \text{ kJ} Second, you need energy to convert the water at 100°C to steam at 100°C. Convert 25.0 g to moles: 25.0 g18.02 g/mol=1.39 mol\frac{25.0 \text{ g}}{18.02 \text{ g/mol}} = 1.39 \text{ mol} Then calculate the vaporization energy: q2=(1.39 mol)(40.7 kJ/mol)=56.6 kJq_2 = (1.39 \text{ mol})(40.7 \text{ kJ/mol}) = 56.6 \text{ kJ} Total heat required: qtotal=2.61+56.6=59.2 kJq_{total} = 2.61 + 56.6 = 59.2 \text{ kJ} This matches answer A (58.9 kJ) within rounding differences. Answer B (54.3 kJ) likely omits the heating step and uses an incorrect mass-to-mole conversion. Answer C (47.1 kJ) probably includes heating but miscalculates the vaporization step. Answer D (40.7 kJ) is simply the molar heat of vaporization given in the problem—a trap for students who forget about the heating step entirely. Study tip: Always break phase change problems into steps: heating to transition temperature, then the phase change itself. Both require energy and must be calculated separately, then added together.

Question 17

A bomb calorimeter with heat capacity 9.15 kJ/°C is used to determine the heat of combustion of a 1.85 g sample of glucose (C6H12O6C_6H_{12}O_6). The temperature increases from 23.42°C to 26.78°C. What is the molar heat of combustion of glucose?

  1. −2990-2990 kJ/mol (correct answer)
  2. −2820-2820 kJ/mol
  3. −2560-2560 kJ/mol
  4. −1860-1860 kJ/mol
  5. −1280-1280 kJ/mol
Explanation: Bomb calorimetry problems test your ability to connect heat transfer with chemical reactions. When you see a calorimeter with a given heat capacity and temperature change, you're calculating the energy released per mole of substance. Start by finding the total heat released using q=C×ΔTq = C \times \Delta T, where CC is the calorimeter's heat capacity and ΔT\Delta T is the temperature change. Here: q=9.15 kJ/°C×(26.78−23.42)°C=9.15×3.36=30.74 kJq = 9.15 \text{ kJ/°C} \times (26.78 - 23.42)°C = 9.15 \times 3.36 = 30.74 \text{ kJ} Next, convert the glucose mass to moles. The molar mass of C6H12O6C_6H_{12}O_6 is 6(12.01)+12(1.008)+6(16.00)=180.16 g/mol6(12.01) + 12(1.008) + 6(16.00) = 180.16 \text{ g/mol}. So: 1.85 g÷180.16 g/mol=0.0103 mol1.85 \text{ g} \div 180.16 \text{ g/mol} = 0.0103 \text{ mol} The molar heat of combustion is: −30.74 kJ0.0103 mol=−2990 kJ/mol\frac{-30.74 \text{ kJ}}{0.0103 \text{ mol}} = -2990 \text{ kJ/mol} (negative because combustion releases energy). Answer A (-2990 kJ/mol) is correct. Answer B (-2820 kJ/mol) likely results from using an incorrect molar mass or rounding errors. Answer C (-2560 kJ/mol) suggests a calculation error in the heat released, possibly using the wrong temperature values. Answer D (-1860 kJ/mol) is too low and might come from confusing the heat capacity units or making errors in unit conversion. Remember: always double-check your molar mass calculation and ensure you're using the correct temperature difference. The sign should be negative for combustion reactions since they release heat.

Question 18

Using the data below, calculate ΔHf°\Delta H_f° for NH3(g)NH_3(g): N2(g)+3H2(g)→2NH3(g)N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) ΔH°=−92.2\Delta H° = -92.2 kJ ΔHf°[N2(g)]=0\Delta H_f°[N_2(g)] = 0 kJ/mol ΔHf°[H2(g)]=0\Delta H_f°[H_2(g)] = 0 kJ/mol

  1. −184.4-184.4 kJ/mol
  2. −92.2-92.2 kJ/mol
  3. −46.1-46.1 kJ/mol (correct answer)
  4. +46.1+46.1 kJ/mol
  5. +92.2+92.2 kJ/mol
Explanation: When you encounter enthalpy of formation problems, remember that ΔHf°\Delta H_f° represents the energy change when one mole of a compound forms from its elements in their standard states. The key is paying attention to stoichiometry. The given reaction shows 2 moles of NH3(g)NH_3(g) forming with ΔH°=−92.2\Delta H° = -92.2 kJ. To find ΔHf°\Delta H_f° for NH3(g)NH_3(g), you need the energy change for forming just 1 mole. Using the relationship ΔHreaction°=∑ΔHf°(products)−∑ΔHf°(reactants)\Delta H_{reaction}° = \sum \Delta H_f°(products) - \sum \Delta H_f°(reactants): −92.2=[2×ΔHf°(NH3)]−[1×ΔHf°(N2)+3×ΔHf°(H2)]-92.2 = [2 \times \Delta H_f°(NH_3)] - [1 \times \Delta H_f°(N_2) + 3 \times \Delta H_f°(H_2)] Since elements in their standard states have ΔHf°=0\Delta H_f° = 0: −92.2=2×ΔHf°(NH3)−0-92.2 = 2 \times \Delta H_f°(NH_3) - 0 Therefore: ΔHf°(NH3)=−92.22=−46.1\Delta H_f°(NH_3) = \frac{-92.2}{2} = -46.1 kJ/mol Choice C is correct. Choice A (-184.4 kJ/mol) incorrectly doubles the total reaction enthalpy instead of dividing by 2. Choice B (-92.2 kJ/mol) uses the total reaction enthalpy without accounting for the 2-mole stoichiometry of NH3NH_3. Choice D (+46.1 kJ/mol) has the right magnitude but wrong sign—formation of ammonia releases energy, so ΔHf°\Delta H_f° must be negative. Study tip: Always check the stoichiometric coefficients in formation reactions. If more than one mole of product forms, divide the total ΔH°\Delta H° by the number of moles to get the per-mole formation enthalpy.