Lesson test: Entropy and free energy

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Entropy and free energy

1.[2p]
Why does the Gibbs free energy criterion replace the requirement that the total entropy rise?

It says the same thing using only properties of the system, which is what an experimenter can measure

Correct

It is an approximation that is easier to compute

The answer is: It says the same thing using only properties of the system, which is what an experimenter can measure

It applies when the surroundings are not at constant temperature

The answer is: It says the same thing using only properties of the system, which is what an experimenter can measure

Total entropy is not conserved, so it cannot be used as a criterion

The answer is: It says the same thing using only properties of the system, which is what an experimenter can measure

2.[3p]
For $2 (g) + (g) \to 2 O(l)$ , $Δ_{r} H^{ominus} = - 571.6$ kJ mol⁻¹ and $Δ_{r} S^{ominus} = - 326.7$ J mol⁻¹ K⁻¹. What is $Δ_{r} G^{ominus}$ at $298.15$ K, in kJ mol⁻¹?

CorrectNot quite: -474.2

3.[3p]
$CaC(s) \to CaO(s) + C(g)$ has $Δ_{r} H^{ominus} = + 178.3$ kJ mol⁻¹ and $Δ_{r} S^{ominus} = + 160.6$ J mol⁻¹ K⁻¹. Above roughly what temperature, in kelvin, does it become favourable at standard conditions?

CorrectNot quite: 1110

4.[1p]
A standard molar entropy can be zero for an element in its standard state, in the way a formation enthalpy is.

True

The answer is: False

False

Correct

5.[2p]
Burning methane has $Δ_{r} S^{ominus} = - 243$ J mol⁻¹ K⁻¹ yet is spontaneous. What resolves this?

The heat released raises the entropy of the surroundings by far more than 243 J per kelvin

Correct

The Second Law does not apply to combustion

The answer is: The heat released raises the entropy of the surroundings by far more than 243 J per kelvin

The entropy of the system is not really negative once the flame is included

The answer is: The heat released raises the entropy of the surroundings by far more than 243 J per kelvin

Enthalpy overrides entropy in exothermic reactions

The answer is: The heat released raises the entropy of the surroundings by far more than 243 J per kelvin

6.[2p]
How much does the entropy of the surroundings change, in J K⁻¹, when a reaction releases $851.5$ kJ at $298.15$ K?

CorrectNot quite: 2856

7.[3p]
Which statements about standard molar entropies are correct?

Select all that apply

Gases have much larger values than the corresponding liquids or solids

Correct

Larger, floppier molecules have larger values than small stiff ones

Correct

Diamond has a very small value because its bonds are stiff

Correct

The value of an element in its standard state is defined as zero

The answer is: Gases have much larger values than the corresponding liquids or solids, Larger, floppier molecules have larger values than small stiff ones, Diamond has a very small value because its bonds are stiff

8.[3p]
Match each sign combination to when the reaction is spontaneous.

Negative H, positive S
Positive H, negative S
Positive H, positive S
Negative H, negative S

only when hot enough
at no temperature
at all temperatures
only when cold enough

Show the answer

Negative H, positive S: at all temperatures Positive H, negative S: at no temperature Positive H, positive S: only when hot enough Negative H, negative S: only when cold enough

9.[2p]
Solid carbon monoxide has a residual entropy of about $4.6$ J mol⁻¹ K⁻¹ at absolute zero. What does this show?

The Third Law applies to perfect crystals, and a frozen-in orientational disorder is an imperfection

Correct

The Third Law is wrong

The answer is: The Third Law applies to perfect crystals, and a frozen-in orientational disorder is an imperfection

Carbon monoxide is not a crystal at low temperature

The answer is: The Third Law applies to perfect crystals, and a frozen-in orientational disorder is an imperfection

Entropy can be negative in some substances

The answer is: The Third Law applies to perfect crystals, and a frozen-in orientational disorder is an imperfection

1.[2p] Why does the Gibbs free energy criterion replace the requirement that the total entropy rise?

2.[3p] For 2(g)+(g)→2O(l), ΔrHominus=-571.6 kJ mol⁻¹ and ΔrSominus=-326.7 J mol⁻¹ K⁻¹. What is ΔrGominus at 298.15 K, in kJ mol⁻¹?

3.[3p] CaC(s)→CaO(s)+C(g) has ΔrHominus=+178.3 kJ mol⁻¹ and ΔrSominus=+160.6 J mol⁻¹ K⁻¹. Above roughly what temperature, in kelvin, does it become favourable at standard conditions?

4.[1p] A standard molar entropy can be zero for an element in its standard state, in the way a formation enthalpy is.

5.[2p] Burning methane has ΔrSominus=-243 J mol⁻¹ K⁻¹ yet is spontaneous. What resolves this?

6.[2p] How much does the entropy of the surroundings change, in J K⁻¹, when a reaction releases 851.5 kJ at 298.15 K?

7.[3p] Which statements about standard molar entropies are correct?

8.[3p] Match each sign combination to when the reaction is spontaneous.

9.[2p] Solid carbon monoxide has a residual entropy of about 4.6 J mol⁻¹ K⁻¹ at absolute zero. What does this show?

1.[2p]
Why does the Gibbs free energy criterion replace the requirement that the total entropy rise?

2.[3p]
For $2 (g) + (g) \to 2 O(l)$ , $Δ_{r} H^{ominus} = - 571.6$ kJ mol⁻¹ and $Δ_{r} S^{ominus} = - 326.7$ J mol⁻¹ K⁻¹. What is $Δ_{r} G^{ominus}$ at $298.15$ K, in kJ mol⁻¹?

3.[3p]
$CaC(s) \to CaO(s) + C(g)$ has $Δ_{r} H^{ominus} = + 178.3$ kJ mol⁻¹ and $Δ_{r} S^{ominus} = + 160.6$ J mol⁻¹ K⁻¹. Above roughly what temperature, in kelvin, does it become favourable at standard conditions?

4.[1p]
A standard molar entropy can be zero for an element in its standard state, in the way a formation enthalpy is.

5.[2p]
Burning methane has $Δ_{r} S^{ominus} = - 243$ J mol⁻¹ K⁻¹ yet is spontaneous. What resolves this?

6.[2p]
How much does the entropy of the surroundings change, in J K⁻¹, when a reaction releases $851.5$ kJ at $298.15$ K?

7.[3p]
Which statements about standard molar entropies are correct?

8.[3p]
Match each sign combination to when the reaction is spontaneous.

9.[2p]
Solid carbon monoxide has a residual entropy of about $4.6$ J mol⁻¹ K⁻¹ at absolute zero. What does this show?