Thermodynamics and Phases - GRE

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Question

$H_{2}+\frac{1}{2}O_{2}\rightarrow H_{2} O\ \ \Delta H_{f}^{0} = - 286 kJ$

$C + O_{2}\rightarrow CO_{2} \ \ \Delta H_{f}^{0} = - 394 kJ$

$8C + 9H_{2}\rightarrow C_{8} H_{18}\ \ \Delta H_{f}^{0} = -269 kJ$

Given the enthalpies of formation, what is the enthalpy of combustion of octane in the reaction:

$2C_{8}H_{18}+25O_{2}\rightarrow 16CO_{2} +18H_{2}O$

Answer

The equation for enthalpy of reaction is:

$\Delta H_{reaction}^{0} = \Delta H_{f, products}^{0} - \Delta H_{f, reactants}^{0}$

Given our chemical reaction and the enthalpies of formation, we can find the enthalpy of reaction.

$2C_{8}H_{18}+25O_{2}\rightarrow 16CO_{2} +18H_{2}O$

First, find the total enthalpy for the products.

$\Delta H_{f, products}^{0} = (16 ) \Delta H_{f, CO_{2}} + (18 ) \Delta H_{f, H_{2}O}$

$\Delta H_{f, products}^{0} = (16 ) (-394kJ) + (18 )(-286kJ) = -11452 kJ$

Then, find the total enthalpy for the reactants.

$\Delta H_{f, reactants}^{0} = (2) \Delta H_{f, C_{8}H_{18}} + (25 ) \Delta H_{f, O_{2}}$

$\Delta H_{f, reactants}^{0} =( 2) (269kJ) + (25) (0kJ) = 538kJ$

Since the oxygen is elemental, its heat of formation is zero.

Return to the original equation to calculate the final enthalpy of reaction.

$\Delta H_{reaction}^{0}=-11452kJ-(538kJ)=-11990kJ\approx-1.2*10^{4}kJ$

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Question

Propane gas combusts according the following chemical equation:

$C_{3}H_{8} + 5O_{2} \rightarrow 3CO_{2} + 4H_{2}O$

Given the following standard enthalpy values, what is the enthalpy of the reaction for the combustion of one mole of propane?

$\Delta H_{C_{3}H_{8}} = -103.8\frac{kJ}{mol}$

$\Delta H_{O_{2}} = 0\frac{kJ}{mol}$

$\Delta H_{CO_{2}} = -393.5\frac{kJ}{mol}$

$\Delta H_{H_{2}O} = -241.8\frac{kJ}{mol}$

Answer

Given the molar enthalpy values for reactants and products, we can solve for the enthalpy of the reaction using the equation:

$\Delta H_{reaction} = \Delta H_{products} - \Delta H_{reactants}$

Keep in mind that the molar enthalpy values must be multiplied by the coefficients that are present in the balanced reaction:

$C_{3}H_{8} + 5O_{2} \rightarrow 3CO_{2} + 4H_{2}O$

$\Delta H_{reaction} = (3\Delta H_{CO_{2}} + 4\Delta H_{H_{2}O} ) - (\Delta H_{C_{3}H_{8}})$

Oxygen is omitted because its enthalpy value is zero.

$\Delta H_{reaction} = (3(-393.5\frac{kJ}{mol})+ 4(-241.8\frac{kJ}{mol})) - (-103.8\frac{kJ}{mol})$

$\Delta H_{reaction} = -2044kJ$

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Question

Which is not characteristic of an endothermic reaction?

Answer

An endothermic reaction absorbs energy in the form of heat. An endothermic reaction involves breaking of chemical bonds because it involves the absorption of energy. Endothermic reactions tend to feel cold because it is taking heat away from your skin. Since endothermic reactions involve absorbing energy, often in the form of heat, the change in enthalpy is positive.

Therefore, the answer is "releasing of energy to surroundings" which is not a characteristic of an endothermic process. Instead, it is a characteristic of an exothermic reaction which is the direct opposite of an endothermic reaction.

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Question

Which of the following reactions has a positive value for change in entropy $(\Delta S > 0)$ ?

Answer

Entropy is a measure of the disorder of a system. In other words, when the components of a system become more uniform and spread out, the entropy of the system has increased. For example, when a liquid becomes gaseous, the molecules separate from one another, increasing the disorder of the system. As a result, the evaporation of water results in an increase in entropy.

All other options result in a more ordered system, which decreases entropy.

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Question

At $25^{\circ}C$ , a reaction has a Gibb's free energy change of . If the enthalpy change of the reaction is , what is the entropy change of the reaction?

Answer

We can relate enthalpy, entropy, and Gibb's free energy using the Gibb's free energy equation:

$\Delta G = \Delta H-T\Delta S$

Keep in mind that temperature must be in Kelvin. Since we know enthalpy and Gibb's free energy for the reaction, we can solve for the entropy change:

$45,000J = 35,000J - (298K)(\Delta S)$

$\Delta S = -33.6\frac{J}{K}$

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Question

Which of the following options indicates that a chemical reaction is unfavorable?

Answer

The sign of $\Delta G$ indicates the direction of a chemical reaction and determines if a reaction is spontaneous or not. A chemical reaction in which $\Delta G< 0$ is considered to be a spontaneous reaction because it will proceed without requiring any outside energy. To approach this problem, we must consider the equation for the Gibbs free energy of a reaction, $\Delta G= \Delta H -T \Delta S$ . A reaction is considered to be favorable when $\Delta G$ is negative. The second law of thermodynamics indicates that the entropy of the universe is spontaneously increasing. This corresponds to a positive $\Delta S$ .

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Question

Use the following values for water as needed.

$\Delta H^o_{fus}=334\frac{J}{g}$

$\Delta H^o_{vap}=2260\frac{J}{g}$

$c=4.186\frac{J}{g^oC}$

$\rho = 1\frac{g}{mL}$

If burning wood releases $\small 20kJ$ of heat energy per gram of wood consumed, what mass of wood must be consumed to heat $\small 1L$ of water from $\small 20^oC$ to $\small 100^oC$ , and then to convert it to water vapor?

Answer

There are two processes requiring added heat in this problem:

1. Raising the temperature of the liquid water from $\small 20^oC$ to $\small 100^oC$ (use $\Delta Q = mc\Delta T$ )

2. Boiling the water at a constant temperature of $\small 100^oC$ (use $\Delta Q = m\Delta H^o_{vap}$ )

To use either of these equation, we need to find the mass of the water using the relation between mass, density, and volume.

$m = \rhoV= 1\frac{g}{mL}1000mL=1000g$

Use this mass with the given specific heat and temperatures to find the heat for part 1 of the process.

$\Delta Q = (1000g)(4.186\frac{J}{g^oC})(100^oC - 20^oC)=334,880J$

Then, use the mass with the given heat of vaporization to find the energy needed to convert the water to water vapor.

$\Delta Q =(1000g) (2260\frac{J}{g})= 2,260,000J$

Sum the energies for step 1 and step 2.

$334,880 J + 2,260,000 J = 2,594,880J\approx 2,595kJ$

This is the total amount of energy needed from the burning wood. Use stoichiometry to find the grams of wood needed to produce this amount of energy.

$2595kJ*(\frac{1 g}{20kJ})=129.7g\approx 130g$

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Question

Calculate the specific heat of of metal that requires of heat energy to raise its temperature from $20.46^{o}C$ to $64.4^{o}C$ ?

Answer

Specific heat is the amount of heat needed to raise 1 gram of a substance by 1oC. Calorimeters used for these types of experiments because they are designed to be well-insulated, so no heat is gained from or lost to the surroundings.

$Specific\ heat(C)=\frac{amount\ of\ heat\ transferred(Q)}{(grams\ of\ substance,m)\times (temperature\ change\bigtriangleup, T)}$

$C=\frac{Q}{m\times \bigtriangleup T}$

$C=\frac{Q}{m\bigtriangleup T}=\frac{410.4J}{(20.4g\times (64.4^{o}C-25.4^{o}C))}=0.516\frac{J}{g^{o}C}$

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Question

What was the final temperature of water if a sample of water absorbs of heat energy and heats up from $25^{o}C$ ? (specific heat of water is $4.18 \frac{J}{g^{o}C}$ )

Answer

Specific heat is the amount of heat needed to raise 1 gram of a substance by 1oC. Calorimeters used for these types of experiments because they are designed to be well-insulated, so no heat is gained from or lost to the surroundings.

$Specific\ heat(C)=\frac{amount\ of\ heat\ transferred(Q)}{(grams\ of\ substance,m)\times (temperature\ change\bigtriangleup, T)}$

$C=\frac{Q}{m\times \bigtriangleup T}$

$Q=mC\bigtriangleup T$

$Q=mC(T_{f}-T_{i})$

$810.5J=19g\times4.179\frac{J}{g^{o}C}\times(T_{f}-25^{o}C)$

$810.5J=79.40\frac{J}{^{o}C}\times(T_{f}-25^{o}C)$

$\frac{810.5J}{79.40\frac{J}{^{o}C}}=T_{f}-25^{o}C$

$10.21^{o}C=T_{f}-25^{o}C$

$T_{f}=35^{o}C$

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Question

An metal at $54.2^{o}C$ was put into of water at $25^{o}C$ . The final temp of the water and metal was $29.3^{o}C$ . Assuming heat was not lost to the surroundings find the specific heat of the metal? The specific heat of water is $4.18\frac{J}{g^{o}C}$ .

Answer

Specific heat is the amount of heat needed to raise 1 gram of a substance by 1oC. Calorimeters used for these types of experiments because they are designed to be well-insulated, so no heat is gained from or lost to the surroundings.

$Specific\ heat(C)=\frac{amount\ of\ heat\ transferred(Q)}{(grams\ of\ substance,m)\times (temperature\ change\bigtriangleup, T)}$

$C=\frac{Q}{m\times \bigtriangleup T}$

Heat is transferred from the metal to the water.

$-Q_{metal}=Q_{water}$

$-(mC\bigtriangleup T)=mC\bigtriangleup T$

$-(mC(T_{f}-T_{i}))=-(mC(T_{f}-T_{i}))$

$-(81.9g\cdot C\cdot (29.3^{o}C-54.2^oC)=40.7g\cdot4.18\frac{J}{g^o}C(29.3^{o}C-25^{o}C)$

$-(-2039g^{o}C\cdot C)=731J$

$2039g^{o}C\cdot\ C=731J$

$C=\frac{731J}{2039g^{o}C}=0.358\frac{J}{g^{o}C}$

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Question

Calculate the final temperature when a sample of metal (specific heat of the metal= $0.52 \frac{J}{g^{o}C}$ ) at $45.0^{o}C$ is placed into of water at $23.0^{o}C$ . (specific heat of water is $4.18\frac{J}{g^{o}C}$ )

Answer

Specific heat is the amount of heat needed to raise 1 gram of a substance by 1oC. Calorimeters used for these types of experiments because they are designed to be well-insulated, so no heat is gained from or lost to the surroundings.

$Specific\ heat(C)=\frac{amount\ of\ heat\ transferred(Q)}{(grams\ of\ substance,m)\times (temperature\ change\bigtriangleup, T)}$

$C=\frac{Q}{m\times \bigtriangleup T}$

Heat is transferred from the metal to the water.

$-Q_{metal}=Q_{water}$

$-(mC\bigtriangleup T)=mC\bigtriangleup T$

$-(mC(T_{f}-T_{i}))=-(mC(T_{f}-T_{i}))$

$-(44.0g\cdot 0.52\frac{J}{g^{o}C}\cdot (T_{f}-45.0^oC)=95.0g\cdot4.18\frac{J}{g^o}C(T_{f}-23^{o}C)$

$-(22.9\frac{J}{^{o}C}(T_{f}-45.0^{o}C))=397\frac{J}{^{o}C}(T_{f}-23^{o}C)$

$-(22.9\frac{J}{^{o}C}\cdot T_{f}-22.9\frac{J}{^{o}C}\cdot 45.0^{o}C)=397\frac{J}{^{o}C}\cdot T_{f}-397\frac{J}{^{o}C}\cdot 23.0^{o}C$

$-(22.9\frac{J}{^{o}C}\cdot T_{f}-1031J)=397\frac{J}{^{o}C}\cdot T_{f}-9131J$

$-22.9\frac{J}{^{o}C}\cdot \ T_{f}+1031J=397\frac{J}{^{o}C}\cdot T_{f}-9131J$

$-22.9\frac{J}{^{o}C}\cdot \ T_{f}-397\frac{J}{^{o}C}\cdot T_{f}=-1031J-9131J$

$-419\frac{J}{^{o}C}\cdot T_{f}=-10162J$

$T_{f}=\frac{-10162^{o}C}{-419.9}=24^{o}C$

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Question

How much heat is absorbed when of water goes from $25.0^{o}C$ to $35.0^{o}C$ ? (Specific heat of water= $4.18 \frac{J}{g^{o}C}$ )

Answer

Specific heat is the amount of heat needed to raise 1 gram of a substance by 1oC. Calorimeters used for these types of experiments because they are designed to be well-insulated, so no heat is gained from or lost to the surroundings.

$Specific\ heat(C)=\frac{amount\ of\ heat\ transferred(Q)}{(grams\ of\ substance,m)\times (temperature\ change\bigtriangleup, T)}$

$C=\frac{Q}{m\times \bigtriangleup T}$

$Q=mC\bigtriangleup T$

$Q=mC(T_{f}-T_{i})$

$Q=150g\cdot\ 4.18\frac{J}{g^{o}C}\cdot (35^{o}C-25^{o}C)$

$Q=6270\ J$

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Question

Calculate the specific heat of of metal that requires of heat energy to raise its temperature from $17.0^{o}C$ to $44.0^{o}C$ ?

Answer

Specific heat is the amount of heat needed to raise 1 gram of a substance by 1oC. Calorimeters used for these types of experiments because they are designed to be well-insulated, so no heat is gained from or lost to the surroundings.

$Specific\ heat(C)=\frac{amount\ of\ heat\ transferred(Q)}{(grams\ of\ substance,m)\times (temperature\ change\bigtriangleup, T)}$

$C=\frac{Q}{m\times \bigtriangleup T}$

$C=\frac{Q}{m\bigtriangleup T}=\frac{353J}{(19.3g\times (44.0^{o}C-17.0^{o}C))}=0.677\frac{J}{g^{o}C}$

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Question

What was the final temperature of the water if a sample of water absorbs of heat energy and heats up from $23^{o}C$ ? (specific heat of water is $4.18 \frac{J}{g^{o}C}$ )

Answer

Specific heat is the amount of heat needed to raise 1 gram of a substance by 1oC. Calorimeters used for these types of experiments because they are designed to be well-insulated, so no heat is gained from or lost to the surroundings.

$Specific\ heat(C)=\frac{amount\ of\ heat\ transferred(Q)}{(grams\ of\ substance,m)\times (temperature\ change\bigtriangleup, T)}$

$C=\frac{Q}{m\times \bigtriangleup T}$

Rearranging gives,

$Q=mC\bigtriangleup T$

$Q=mC(T_{f}-T_{i})$

$423J=19g\times4.18\frac{J}{g^{o}C}\times(T_{f}-23^{o}C)$

$423J=79\frac{J}{^{o}C}\times(T_{f}-23^{o}C)$

$\frac{423J}{79\frac{J}{^{o}C}}=T_{f}-23^{o}C$

$5.35^{o}C=T_{f}-23^{o}C$

$T_{f}=28^{o}C$

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Question

A metal at $37.1^{o}C$ was placed in of water at $24.2^{o}C$ . The final temp of the water and metal was $26.0^{o}C$ . Assuming heat was not lost to the surroundings find the specific heat of the metal? The specific heat of water is $4.18\frac{J}{g^{o}C}$ .

Answer

Specific heat is the amount of heat needed to raise 1 gram of a substance by 1oC. Calorimeters used for these types of experiments because they are designed to be well-insulated, so no heat is gained from or lost to the surroundings.

$Specific\ heat(C)=\frac{amount\ of\ heat\ transferred(Q)}{(grams\ of\ substance,m)\times (temperature\ change\bigtriangleup, T)}$

$C=\frac{Q}{m\times \bigtriangleup T}$

Heat is transferred from the metal to the water.

$-Q_{metal}=Q_{water}$

$-(mC\bigtriangleup T)=mC\bigtriangleup T$

$-(mC(T_{f}-T_{i}))=-(mC(T_{f}-T_{i}))$

$-(20.5g\cdot C\cdot (26.0^{o}C-37.1^oC)=23.3g\cdot4.18\frac{J}{g^o}C(26.0^{o}C-24.2^{o}C)$

$-(-227.55g^{o}C\cdot C)=175.3092J$

$227.55g^{o}C\cdot\ C=175.3092J$

$C=\frac{175.3092J}{227.55g^{o}C}=0.77\frac{J}{g^{o}C}$

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Question

Calculate the final temperature when a sample of metal (specific heat of the metal= $0.72 \frac{J}{g^{o}C}$ ) at $37.0^{o}C$ is placed into of water at $26.2^{o}C$ . (Specific heat of water is $4.18 \frac{J}{g^{o}C}$ )

Answer

Specific heat is the amount of heat needed to raise 1 gram of a substance by 1oC. Calorimeters used for these types of experiments because they are designed to be well-insulated, so no heat is gained from or lost to the surroundings.

$Specific\ heat(C)=\frac{amount\ of\ heat\ transferred(Q)}{(grams\ of\ substance,m)\times (temperature\ change\bigtriangleup, T)}$

$C=\frac{Q}{m\times \bigtriangleup T}$

Heat is transferred from the metal to the water.

$-Q_{metal}=Q_{water}$

$-(mC\bigtriangleup T)=mC\bigtriangleup T$

$-(mC(T_{f}-T_{i}))=-(mC(T_{f}-T_{i}))$

$-(14.7g\cdot 0.72\frac{J}{g^{o}C}\cdot (T_{f}-37.0^oC)=31.7g\cdot4.18\frac{J}{g^o}C(T_{f}-26.2^{o}C)$

$-(10.584\frac{J}{^{o}C}(T_{f}-37.0^{o}C))=132.51\frac{J}{^{o}C}(T_{f}-26.2^{o}C)$

$-(10.584\frac{J}{^{o}C}\cdot T_{f}-10.584\frac{J}{^{o}C}\cdot 37.0^{o}C)=132.51\frac{J}{^{o}C}\cdot T_{f}-132.51\frac{J}{^{o}C}\cdot 26.2^{o}C$

$-(10.584\frac{J}{^{o}C}\cdot T_{f}-391.61J)=132.51\frac{J}{^{o}C}\cdot T_{f}-3472J$

$-10.584\frac{J}{^{o}C}\cdot T_{f}+391.61J=132.51\frac{J}{^{o}C}\cdot T_{f}-3472J$

$-10.584\frac{J}{^{o}C}\cdot T_{f}-132.51\frac{J}{^{o}C}\cdot T_{f}=-3472J-391.61J$

$-143.094\frac{J}{^{o}C}\cdot T_{f}=-3863.61J$

$T_{f}=\frac{-3863.61^{o}C}{-143.094}=27^{o}C$

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Question

How much heat is absorbed when of water goes from $24.5^{o}C$ to $43.0^{o}C$ ? (Specific heat of water = $4.18 \frac{J}{g^{o}C}$ )

Answer

Specific heat is the amount of heat needed to raise 1 gram of a substance by 1oC. Calorimeters used for these types of experiments because they are designed to be well-insulated, so no heat is gained from or lost to the surroundings.

$Specific\ heat(C)=\frac{amount\ of\ heat\ transferred(Q)}{(grams\ of\ substance,m)\times (temperature\ change\bigtriangleup, T)}$

$C=\frac{Q}{m\times \bigtriangleup T}$

Rearranging gives,

$Q=mC\bigtriangleup T$

$Q=mC(T_{f}-T_{i})$

$Q=50g\cdot\ 4.18\frac{J}{g^{o}C}\cdot (43.0^{o}C-24.5^{o}C)$

$Q=3867\ J$

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Question

What was the final temperature of the water if a sample of water absorbs of heat energy and heats up from $25.0^{o}C$ ? (specific heat of water= $4.18\frac{J}{g^{o}C}$ )

Answer

Specific heat is the amount of heat needed to raise 1 gram of a substance by 1oC. Calorimeters used for these types of experiments because they are designed to be well-insulated, so no heat is gained from or lost to the surroundings.

$Specific\ heat(C)=\frac{amount\ of\ heat\ transferred(Q)}{(grams\ of\ substance,m)\times (temperature\ change\bigtriangleup, T)}$

$C=\frac{Q}{m\times \bigtriangleup T}$

Rearranging gives,

$Q=mC\bigtriangleup T$

$Q=mC(T_{f}-T_{i})$

$231J=112g\times4.18\frac{J}{g^{o}C}\times(T_{f}-25.0^{o}C)$

$231J=468.16\frac{J}{^{o}C}\times(T_{f}-25.0^{o}C)$

$\frac{231J}{468.16\frac{J}{^{o}C}}=T_{f}-25.0^{o}C$

$0.493^{o}C=T_{f}-25.0^{o}C$

$T_{f}=25.5^{o}C$

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Question

A metal at $72.9^{o}C$ was placed in of water at $25.0^{o}C$ . The final temp of the water and metal was $53.3^{o}C$ . Assuming heat was not lost to the surroundings find the specific heat of the metal? The specific heat of water is $4.18\frac{J}{g^{o}C}$ .

Answer

Specific heat is the amount of heat needed to raise 1 gram of a substance by 1oC. Calorimeters used for these types of experiments because they are designed to be well-insulated, so no heat is gained from or lost to the surroundings.

$Specific\ heat(C)=\frac{amount\ of\ heat\ transferred(Q)}{(grams\ of\ substance,m)\times (temperature\ change\bigtriangleup, T)}$

$C=\frac{Q}{m\times \bigtriangleup T}$

Heat is transferred from the metal to the water.

$-Q_{metal}=Q_{water}$

$-(mC\bigtriangleup T)=mC\bigtriangleup T$

$-(mC(T_{f}-T_{i}))=-(mC(T_{f}-T_{i}))$

$-(10.2g\cdot C\cdot (53.3^{o}C-72.9^oC)=21.2g\cdot4.18\frac{J}{g^o}C(53.3^{o}C-25.0^{o}C)$

$-(-199.92g^{o}C\cdot C)=2507.83J$

$199.92g^{o}C\cdot C=2507.83J$

$C=\frac{2507.83J}{199.92g^{o}C}=12.5\frac{J}{g^{o}C}$

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Question

Which law of thermodynamics states that a crystal's entropy at a temperature of absolute zero is equal to 0?

Answer

There are four laws of thermodynamics:

Zeroth law of thermodynamics: If two independent systems are each in thermodynamic equilibrium with a thrid independent system, then the first two systems must be in thermodyanmic equilibrium with each other.

First law of thermodynamics: Energy is neither created nor destroyed.

Second law of thermodynamics: The entropy of the universe increases following every reaction.

Third law of termodynamics: At a temperature of 0 Kelvin (absolute zero), the entropy of the crystal is 0.

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