A chemical engineer is studying the two reactions shown in the table below. In each case, he fills a reaction vessel with some mixture of the reactants and products at a constant temperature of 58.0 °C and constant total pressure. Then, he measures the reaction enthalpy AH and reaction entropy A.S of the first reaction, and the reaction enthalpy AH and reaction free energy AG of the second reaction. The results of his measurements are shown in the table. Complete the table. That is, calculate AG for the first reaction and AS for the second. (Round your answer to zero decimal places.) Then, decide whether, under the conditions the engineer has set up, the reaction is spontaneous, the reverse reaction is spontaneous, or neither forward nor reverse reaction is spontaneous because the system is at equilibrium. 4PF; (g) + 10H₂(g) → P₂ (s) + 20HF (g) 2NH₂(g) → N₂(g) + 3H₂(g) AH = 1005. kJ J AS 3114. AG = kJ Which is spontaneous? O this reaction O the reverse reaction O neither AH = 92. kJ AS = AG = 0. kJ Which is spontaneous? O this reaction O the reverse reaction O neither X 06 S
A chemical engineer is studying the two reactions shown in the table below. In each case, he fills a reaction vessel with some mixture of the reactants and products at a constant temperature of 58.0 °C and constant total pressure. Then, he measures the reaction enthalpy AH and reaction entropy A.S of the first reaction, and the reaction enthalpy AH and reaction free energy AG of the second reaction. The results of his measurements are shown in the table. Complete the table. That is, calculate AG for the first reaction and AS for the second. (Round your answer to zero decimal places.) Then, decide whether, under the conditions the engineer has set up, the reaction is spontaneous, the reverse reaction is spontaneous, or neither forward nor reverse reaction is spontaneous because the system is at equilibrium. 4PF; (g) + 10H₂(g) → P₂ (s) + 20HF (g) 2NH₂(g) → N₂(g) + 3H₂(g) AH = 1005. kJ J AS 3114. AG = kJ Which is spontaneous? O this reaction O the reverse reaction O neither AH = 92. kJ AS = AG = 0. kJ Which is spontaneous? O this reaction O the reverse reaction O neither X 06 S
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Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![### Chemical Reaction Analysis
#### Engineer's Experimental Setup:
A chemical engineer is analyzing two reactions detailed in the table below. For each reaction, the engineer uses a reaction vessel containing a mixture of reactants and products, maintaining a constant temperature of 58.0 °C and constant total pressure. The engineer measures the reaction enthalpy (ΔH) and entropy (ΔS) of the first reaction, and the reaction enthalpy (ΔH) and reaction free energy (ΔG) of the second reaction. The acquired data are presented below.
Your task is to complete the table by calculating the Gibbs free energy (ΔG) for the first reaction and the entropy change (ΔS) for the second reaction. Subsequently, determine whether the reaction is spontaneous, the reverse reaction is spontaneous, or neither direction is spontaneous under the given conditions.
#### Data and Calculations:
Temperature: 58.0 °C (convert to Kelvin for calculations by adding 273.15, thus 331.15 K).
##### Reaction 1:
\[ 4PF_5(g) + 10H_2(g) \rightarrow P_4(s) + 20HF(g) \]
- ΔH = 1005. kJ
- ΔS = 3114. J/K (Note the unit, convert kJ to J if necessary for consistency)
- ΔG (to be calculated)
Gibbs Free Energy formula:
\[ \Delta G = \Delta H - T\Delta S \]
\[ \Delta G = 1005 \,\text{kJ} - 331.15 \, \text{K} \times \left( \frac{3114 \, \text{J/K}}{1000} \right) \]
\[ \Delta G = 1005 \,\text{kJ} - 331.15 \, \text{K} \times 3.114 \, \text{kJ/K} \]
\[ \Delta G = 1005 \,\text{kJ} - 1031.76 \, \text{kJ} \]
\[ \Delta G = -26.76 \,\text{kJ} \]
Finalize to zero decimal places:
\[ \Delta G \approx -27 \, \text{kJ} \]
**Spontaneity:**
- Since ΔG < 0, "this reaction" (the forward reaction) is spontaneous.
##### Reaction](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F12c0a0e9-4d98-45bb-8c42-bb6fd08d9097%2F80f35b19-2774-41bc-8b13-c7ec9e959e25%2Felo1wz_processed.png&w=3840&q=75)
Transcribed Image Text:### Chemical Reaction Analysis
#### Engineer's Experimental Setup:
A chemical engineer is analyzing two reactions detailed in the table below. For each reaction, the engineer uses a reaction vessel containing a mixture of reactants and products, maintaining a constant temperature of 58.0 °C and constant total pressure. The engineer measures the reaction enthalpy (ΔH) and entropy (ΔS) of the first reaction, and the reaction enthalpy (ΔH) and reaction free energy (ΔG) of the second reaction. The acquired data are presented below.
Your task is to complete the table by calculating the Gibbs free energy (ΔG) for the first reaction and the entropy change (ΔS) for the second reaction. Subsequently, determine whether the reaction is spontaneous, the reverse reaction is spontaneous, or neither direction is spontaneous under the given conditions.
#### Data and Calculations:
Temperature: 58.0 °C (convert to Kelvin for calculations by adding 273.15, thus 331.15 K).
##### Reaction 1:
\[ 4PF_5(g) + 10H_2(g) \rightarrow P_4(s) + 20HF(g) \]
- ΔH = 1005. kJ
- ΔS = 3114. J/K (Note the unit, convert kJ to J if necessary for consistency)
- ΔG (to be calculated)
Gibbs Free Energy formula:
\[ \Delta G = \Delta H - T\Delta S \]
\[ \Delta G = 1005 \,\text{kJ} - 331.15 \, \text{K} \times \left( \frac{3114 \, \text{J/K}}{1000} \right) \]
\[ \Delta G = 1005 \,\text{kJ} - 331.15 \, \text{K} \times 3.114 \, \text{kJ/K} \]
\[ \Delta G = 1005 \,\text{kJ} - 1031.76 \, \text{kJ} \]
\[ \Delta G = -26.76 \,\text{kJ} \]
Finalize to zero decimal places:
\[ \Delta G \approx -27 \, \text{kJ} \]
**Spontaneity:**
- Since ΔG < 0, "this reaction" (the forward reaction) is spontaneous.
##### Reaction
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