AP Chemistry : Thermochemistry and Kinetics

Study concepts, example questions & explanations for AP Chemistry

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Example Questions

Example Question #17 : Reaction Rate And Rate Law

  (slow)

  (fast)

The mechanism for decomposition of ozone is shown. What is the intermediate of the process?

Possible Answers:

Correct answer:

Explanation:

Intermediate is created and destroyed, and therefore does not appear on the net equation, which is . Thus, the intermediate is . Note that when asked for an intermediate, the coefficient in front of it is not used, rather we are looking for the species that is a product of one reaction and a reactant in a subsequent step.

Example Question #18 : Reaction Rate And Rate Law

The rate law of the reaction, , is . Which of the following does not increase the rate of the reaction?

Possible Answers:

Increasing the concentration of 

Adding the catalyst to the reaction

Increasing the temperature of the reaction

Increasing the concentration of 

Correct answer:

Increasing the concentration of 

Explanation:

The reactant  is not included in the rate law expression, and therefore altering its concentration does not affect the rate of the reaction. Catalysts always increase the rate of reactions by lowering its activation energy. Increasing temperature (average kinetic energy of the molecules) increases the frequency of collisions, and increases the proportion of collisions that have enough energy to overcome the activation energy and undergo a chemical reaction. Increasing the concentration of  will increase the rate of the reaction as indicated by the rate law.

Example Question #19 : Reaction Rate And Rate Law

Which of the following changes to reaction conditions will always result in an increase in the reaction rate?

Possible Answers:

Increased concentration of reactants

Increased temperature

Decreased temperature

Decreased concentration of products

Correct answer:

Increased temperature

Explanation:

For this question, we're asked to identify something that will always increase the rate of a reaction. Notice that the question specifically states "always."

Let's go through each answer choice and see how it affects the reaction rate.

When the concentration of reactants is increased, this may increase the reaction rate, but not always. For example, if a reaction is zero-order with respect to its reactants, then changing the reactant concentration will have no effect on the rate.

Just like with reactants, decreasing the concentration of products may or may not change the reaction rate. If the reaction rate is zero-order with respect to the products, then a change in their concentration will have no effect on the reaction rate.

A change in temperature is the only thing that is guaranteed to change the reaction rate. This is because changing the temperature will directly change the rate constant of the reaction. Increasing the temperature will increase the rate constant, and hence the reaction rate.

Example Question #1 : Calorimetry, Specific Heat, And Calculations

The following is a list of specific heat capacities for a few metals.

A 50g sample of an unknown metal is heated with 800 joules. If the temperature of the metal increases by 41.6oC, what is the identity of the unknown metal?

Possible Answers:

Aluminum

Copper

Iron

Silver

Correct answer:

Copper

Explanation:

We need to find the specific heat of the unknown sample of metal in order to locate it on the list. We can do this by using the equation that allows us to determine the specific heat capacity of an element.

Since we know the change in temperature, we can simply plug in the values and solve for the value of .

Going back to the list, we see that this is the specific heat capacity for copper, so we confirm that the unknown metal is copper. 

Example Question #2 : Calorimetry, Specific Heat, And Calculations

In which instance would a bomb calorimeter be more useful than a coffee-cup calorimeter?

Possible Answers:

When measuring the mass of a substance

When measuring the specific heat of water vapor

When measuring the heat capacity of a solid

None of the above

Correct answer:

When measuring the specific heat of water vapor

Explanation:

Bomb calorimeters are most useful when dealing with a gas, because they can operate well at high pressures. Coffee-cup calorimeters are not useful when water begins to boil, producing vapor.

Example Question #3 : Calorimetry, Specific Heat, And Calculations

How much heat does it take to heat 100g ice at 0C to boiling point?

Cice= 2.1 J/goC

Cwater= 4.2 J/goC

ΔHvap= 2260 J/g

ΔHfus=334 J/g

Possible Answers:

42 kJ

75.4 kJ

55.4 kJ

300.5 kJ

33.4 kJ

Correct answer:

75.4 kJ

Explanation:

You need heat for the phase change, using the enthalpy of fusion (100g*334 J/g = 33400 J). Add to this the heat to get to boiling point using the specific heat of water (100g*100C*4.2 J/goC = 42000 J). Totalling 75400 J (75.4 kJ)

Example Question #4 : Calorimetry, Specific Heat, And Calculations

The specific heat capacity of an unknown liquid is 0.32\frac{J}{kg\cdot K}. The density of the liquid is 0.0321 \frac{g}{mL} If a chemist applies 243 J of heat to 300 mL of this liquid starting at 27.1^{\circ}C, what is the final temperature?

Possible Answers:

78882^{\circ}C

None of the available answers

2700.1^{\circ}C

270.1^{\circ}C

78882\hspace{1 mm}K

Correct answer:

78882^{\circ}C

Explanation:

First we will determine the mass of the liquid:

300\hspace{1 mm}mL\times\frac{0.0321\hspace{1 mm}g}{1\hspace{1 mm}mL}=9.63\hspace{1 mm}g

Now we will examine the relationship between heat and specific heat capacity:

Q=cm\Delta T

Where Q is heat in Joules, c is the specific heat capacity, m is the mass and \Delta T is the change in temperature. We can rearrange this

\Delta T=\frac{Q}{cm}

\Delta T=\frac{243\hspace{1 mm}J}{0.32J\cdot kg^{-1}\cdot K^{-1}0.00963kg}=78855\hspace{1 mm}K

If we begin at 27.1^{\circ}C, we will end at 78882^{\circ}C

 

Example Question #5 : Calorimetry, Specific Heat, And Calculations

A 50g sample of a metal was heated to  then quickly transferred to an insulated container containing 50g of  at .  The final temperature of the  was .

Which of the following can be concluded?

Possible Answers:

None of the other answers

The specific heat of the metal is greater than that of the water

The water gained an amount of thermal energy that was more than the amount of thermal energy lost by the metal

The specific heat of the water is greater than that of the metal

The metal lost an amount of thermal energy that was more than the amount of thermal energy gained by the water

Correct answer:

The specific heat of the water is greater than that of the metal

Explanation:

When the heated metal is placed in the container of the cooler water there will be a transfer of thermal energy from the metal to the water.  This transfer will occur towards an equilibrium of thermal energy in the water and in the metal. Thus we can conclude that the amount of thermal energy lost by the metal will equal the amount of thermal energy gained by the water.  However we notice that the water increases by only 5oC  and the metal decreases by 65oC.  This is becasue of the difference of the specific heats of these substances.  The specific heat capacity of a substance is the heat required to increase the temperature of 1g of a substance by 1oC. The metal can be conluded to have a smaller specific heat than the water because the same amount of energy transfer led to a much larger change in termperature for the metal as compared to the water.  

Example Question #6 : Calorimetry, Specific Heat, And Calculations

Which of the following is the correct molar specific heat of water used when making calculations involving a calorimeter?

Possible Answers:

4.184 J/gK

418.4 J/gK

41.84 J/gK

0.4184 J/gK

Correct answer:

4.184 J/gK

Explanation:

4.184 J/gK is the cited value for the specific heat of water and should be memorized. This is used during calorimeter calculations, specifically when using the equation q= mc delta(T).

Example Question #7 : Calorimetry, Specific Heat, And Calculations

How much energy is needed to raise the temperature of five grams of ice from  to ?

Possible Answers:

Correct answer:

Explanation:

This question involves the total energy needed for three different processes: the temperature raise from  to , the melting of the ice, and the temperature raise from  to . For the first and third transitions we will use the equation . For the melting of ice, we will use the equation .

1. 

2. 

3. 

Finally, we will need to sum the energy required for each step to find the total energy.

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