Free preview 2/16

Question 1
A reaction has $Δ H^{\circ} = + 84 kJ mol^{- 1}$ and $Δ S^{\circ} = + 210 J mol^{- 1} K^{- 1}$ . Below what temperature (in K) is the reaction non-spontaneous?
No clue? Show me the answer
Correct answerCorrect!Incorrect
B400 K
Step-by-step walkthrough
Choose a solution method
Method #1Approach 1
Step 1: Identify the condition for the boundary between spontaneous and non-spontaneous
The reaction switches from non-spontaneous to spontaneous when $Δ G = 0$ . Using $Δ G = Δ H - T Δ S$ , set $Δ G = 0$ to find the threshold temperature.
Step 2: Convert $\Delta S^\circ$ to consistent units
Convert $Δ S^{\circ}$ from J to kJ: $Δ S^{\circ} = \frac{210}{1000} = 0.210 kJ mol^{- 1} K^{- 1}$
Step 3: Solve for the threshold temperature
$T = \frac{Δ H ^{\circ}}{Δ S ^{\circ}} = \frac{84}{0.210} = 400 K$
Step 4: Interpret the result
Since $Δ H > 0$ and $Δ S > 0$ , the reaction is spontaneous only above the threshold temperature. Below 400 K, $Δ G > 0$ and the reaction is non-spontaneous. The correct answer is 400 K.
Method #2Approach 2
Step 1: Identify what is being asked
We need the temperature below which the reaction is non-spontaneous. For a reaction with $Δ H > 0$ and $Δ S > 0$ , the threshold is $T = Δ H /Δ S$ .
Step 2: Eliminate 200 K
$T = 84/0.210 = 400 K$ , not 200 K. 200 K would require $Δ H /Δ S = 200$ , which would mean $Δ S = 0.420 kJ mol^{- 1} K^{- 1}$ , inconsistent with the given data.
Step 3: Eliminate 500 K
$500 K$ does not equal $84/0.210$ . This would be obtained only if $Δ H$ were $+ 105 kJ mol^{- 1}$ , not the given value.
Step 4: Eliminate 840 K
840 K results from forgetting to convert $Δ S$ to kJ (i.e., using $Δ S = 0.100$ instead of $0.210$ ) — a classic unit error. This is incorrect.
Step 5: Select the correct answer
The correct threshold temperature is $T = 84/0.210 = 400 K$ . Below this temperature the reaction is non-spontaneous.
Question 2
Which of the following best explains why the evaporation of ethanol at room temperature is a spontaneous process despite being endothermic?
No clue? Show me the answer
Correct answerCorrect!Incorrect
CThe large positive entropy change of the system makes $T Δ S > Δ H$ , giving $Δ G < 0$
Step-by-step walkthrough
Choose a solution method
Method #1Approach 1
Step 1: Identify the thermodynamic signs for evaporation
Evaporation is endothermic ( $Δ H > 0$ ) and produces gas from liquid, so $Δ S > 0$ (a large increase in entropy as liquid becomes vapour).
Step 2: Apply the Gibbs equation
$Δ G = Δ H - T Δ S$ For $Δ G < 0$ when $Δ H > 0$ , we need $T Δ S > Δ H$ , i.e., the entropy term must outweigh the enthalpy term.
Step 3: Classify this as the $\Delta H > 0$, $\Delta S > 0$ scenario
This corresponds to the case where spontaneity occurs at sufficiently high temperatures. At room temperature, the $T Δ S$ term is large enough (due to the substantial entropy increase from liquid to gas) to make $Δ G < 0$ .
Step 4: Select the correct explanation
The correct answer is that the large positive entropy change makes $T Δ S > Δ H$ , giving $Δ G < 0$ and spontaneity.
Method #2Approach 2
Step 1: Identify what is being asked
We need to explain why an endothermic evaporation is spontaneous. This requires applying $Δ G = Δ H - T Δ S$ .
Step 2: Eliminate 'the enthalpy change is small enough'
This is vague and misleading. Spontaneity is determined by the relationship between $Δ H$ and $T Δ S$ , not just the magnitude of $Δ H$ alone.
Step 3: Eliminate 'the large increase in entropy of the surroundings'
For an endothermic process, the surroundings actually lose heat, which decreases the entropy of the surroundings. This option is factually incorrect.
Step 4: Eliminate 'endothermic processes are always spontaneous at room temperature'
This is clearly false — many endothermic processes are non-spontaneous at room temperature (e.g., the decomposition of water).
Step 5: Select the correct answer
The only correct explanation is that the large positive $Δ S$ means $T Δ S > Δ H$ , so $Δ G < 0$ and the process is spontaneous.

Free preview

14 more questions in this topic

Free preview 2/16

Question 1
A reaction has $Δ H^{\circ} = + 84 kJ mol^{- 1}$ and $Δ S^{\circ} = + 210 J mol^{- 1} K^{- 1}$ . Below what temperature (in K) is the reaction non-spontaneous?
No clue? Show me the answer
Correct answerCorrect!Incorrect
B400 K
Step-by-step walkthrough
Choose a solution method
Method #1Approach 1
Step 1: Identify the condition for the boundary between spontaneous and non-spontaneous
The reaction switches from non-spontaneous to spontaneous when $Δ G = 0$ . Using $Δ G = Δ H - T Δ S$ , set $Δ G = 0$ to find the threshold temperature.
Step 2: Convert $\Delta S^\circ$ to consistent units
Convert $Δ S^{\circ}$ from J to kJ: $Δ S^{\circ} = \frac{210}{1000} = 0.210 kJ mol^{- 1} K^{- 1}$
Step 3: Solve for the threshold temperature
$T = \frac{Δ H ^{\circ}}{Δ S ^{\circ}} = \frac{84}{0.210} = 400 K$
Step 4: Interpret the result
Since $Δ H > 0$ and $Δ S > 0$ , the reaction is spontaneous only above the threshold temperature. Below 400 K, $Δ G > 0$ and the reaction is non-spontaneous. The correct answer is 400 K.
Method #2Approach 2
Step 1: Identify what is being asked
We need the temperature below which the reaction is non-spontaneous. For a reaction with $Δ H > 0$ and $Δ S > 0$ , the threshold is $T = Δ H /Δ S$ .
Step 2: Eliminate 200 K
$T = 84/0.210 = 400 K$ , not 200 K. 200 K would require $Δ H /Δ S = 200$ , which would mean $Δ S = 0.420 kJ mol^{- 1} K^{- 1}$ , inconsistent with the given data.
Step 3: Eliminate 500 K
$500 K$ does not equal $84/0.210$ . This would be obtained only if $Δ H$ were $+ 105 kJ mol^{- 1}$ , not the given value.
Step 4: Eliminate 840 K
840 K results from forgetting to convert $Δ S$ to kJ (i.e., using $Δ S = 0.100$ instead of $0.210$ ) — a classic unit error. This is incorrect.
Step 5: Select the correct answer
The correct threshold temperature is $T = 84/0.210 = 400 K$ . Below this temperature the reaction is non-spontaneous.
Question 2
Which of the following best explains why the evaporation of ethanol at room temperature is a spontaneous process despite being endothermic?
No clue? Show me the answer
Correct answerCorrect!Incorrect
CThe large positive entropy change of the system makes $T Δ S > Δ H$ , giving $Δ G < 0$
Step-by-step walkthrough
Choose a solution method
Method #1Approach 1
Step 1: Identify the thermodynamic signs for evaporation
Evaporation is endothermic ( $Δ H > 0$ ) and produces gas from liquid, so $Δ S > 0$ (a large increase in entropy as liquid becomes vapour).
Step 2: Apply the Gibbs equation
$Δ G = Δ H - T Δ S$ For $Δ G < 0$ when $Δ H > 0$ , we need $T Δ S > Δ H$ , i.e., the entropy term must outweigh the enthalpy term.
Step 3: Classify this as the $\Delta H > 0$, $\Delta S > 0$ scenario
This corresponds to the case where spontaneity occurs at sufficiently high temperatures. At room temperature, the $T Δ S$ term is large enough (due to the substantial entropy increase from liquid to gas) to make $Δ G < 0$ .
Step 4: Select the correct explanation
The correct answer is that the large positive entropy change makes $T Δ S > Δ H$ , giving $Δ G < 0$ and spontaneity.
Method #2Approach 2
Step 1: Identify what is being asked
We need to explain why an endothermic evaporation is spontaneous. This requires applying $Δ G = Δ H - T Δ S$ .
Step 2: Eliminate 'the enthalpy change is small enough'
This is vague and misleading. Spontaneity is determined by the relationship between $Δ H$ and $T Δ S$ , not just the magnitude of $Δ H$ alone.
Step 3: Eliminate 'the large increase in entropy of the surroundings'
For an endothermic process, the surroundings actually lose heat, which decreases the entropy of the surroundings. This option is factually incorrect.
Step 4: Eliminate 'endothermic processes are always spontaneous at room temperature'
This is clearly false — many endothermic processes are non-spontaneous at room temperature (e.g., the decomposition of water).
Step 5: Select the correct answer
The only correct explanation is that the large positive $Δ S$ means $T Δ S > Δ H$ , so $Δ G < 0$ and the process is spontaneous.

Free preview

Step-by-step walkthrough

Step 1: Identify the condition for the boundary between spontaneous and non-spontaneous

Step 2: Convert $\Delta S^\circ$ to consistent units

Step 3: Solve for the threshold temperature

Step 4: Interpret the result

Step 1: Identify what is being asked

Step 2: Eliminate 200 K

Step 3: Eliminate 500 K

Step 4: Eliminate 840 K

Step 5: Select the correct answer

Step-by-step walkthrough

Step 1: Identify the thermodynamic signs for evaporation

Step 2: Apply the Gibbs equation

Step 3: Classify this as the $\Delta H > 0$, $\Delta S > 0$ scenario

Step 4: Select the correct explanation

Step 1: Identify what is being asked

Step 2: Eliminate 'the enthalpy change is small enough'

Step 3: Eliminate 'the large increase in entropy of the surroundings'

Step 4: Eliminate 'endothermic processes are always spontaneous at room temperature'

Step 5: Select the correct answer

14 more questions in this topic

Step-by-step walkthrough

Step 1: Identify the condition for the boundary between spontaneous and non-spontaneous

Step 2: Convert $\Delta S^\circ$ to consistent units

Step 3: Solve for the threshold temperature

Step 4: Interpret the result

Step 1: Identify what is being asked

Step 2: Eliminate 200 K

Step 3: Eliminate 500 K

Step 4: Eliminate 840 K

Step 5: Select the correct answer

Step-by-step walkthrough

Step 1: Identify the thermodynamic signs for evaporation

Step 2: Apply the Gibbs equation

Step 3: Classify this as the $\Delta H > 0$, $\Delta S > 0$ scenario

Step 4: Select the correct explanation

Step 1: Identify what is being asked

Step 2: Eliminate 'the enthalpy change is small enough'

Step 3: Eliminate 'the large increase in entropy of the surroundings'

Step 4: Eliminate 'endothermic processes are always spontaneous at room temperature'

Step 5: Select the correct answer

14 more questions in this topic