QuestionApril 24, 2025

14. Consider the reaction C_(2)H_(4)(g)+H_(2)(g)leftharpoons C_(2)H_(6)(g) where Delta G^circ =-100.4kJ at 25^circ C. Calculate Delta G for this reaction at 25^circ C under the following conditions: P(C_(2)H_(4))=0.211atm P(H_(2))=0.174atm P(C_(2)H_(6))=0.744atm A. -107.9kJ B. -102.0kJ C. -98.8kJ D. -97.2kJ E. -92.9kJ

14. Consider the reaction C_(2)H_(4)(g)+H_(2)(g)leftharpoons C_(2)H_(6)(g) where Delta G^circ =-100.4kJ at 25^circ C. Calculate Delta G for this reaction at 25^circ C under the following conditions: P(C_(2)H_(4))=0.211atm P(H_(2))=0.174atm P(C_(2)H_(6))=0.744atm A. -107.9kJ B. -102.0kJ C. -98.8kJ D. -97.2kJ E. -92.9kJ
14. Consider the reaction C_(2)H_(4)(g)+H_(2)(g)leftharpoons C_(2)H_(6)(g) where Delta G^circ =-100.4kJ at 25^circ C. Calculate Delta G for this reaction at 25^circ C under the following conditions: P(C_(2)H_(4))=0.211atm P(H_(2))=0.174atm P(C_(2)H_(6))=0.744atm A. -107.9kJ B. -102.0kJ C. -98.8kJ D. -97.2kJ E. -92.9kJ

Solution
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Answer

E. -92.9kJ Explanation 1. Calculate Reaction Quotient Q Q = \frac{P(C_{2}H_{6})}{P(C_{2}H_{4}) \cdot P(H_{2})} = \frac{0.744}{0.211 \times 0.174} 2. Compute Q Q = \frac{0.744}{0.036714} = 20.26 3. Apply Gibbs Free Energy Formula \Delta G = \Delta G^{\circ} + RT \ln Q, where R = 8.314 \, J/(mol \cdot K) and T = 298 \, K 4. Convert Units \Delta G^{\circ} = -100.4 \, kJ = -100400 \, J 5. Calculate \Delta G \Delta G = -100400 + (8.314 \times 298 \times \ln(20.26)) 6. Compute \Delta G \Delta G = -100400 + 8.314 \times 298 \times 3.009 \Delta G = -100400 + 7477.9 \Delta G = -92822.1 \, J = -92.8221 \, kJ

Explanation

1. Calculate Reaction Quotient $Q$<br /> $Q = \frac{P(C_{2}H_{6})}{P(C_{2}H_{4}) \cdot P(H_{2})} = \frac{0.744}{0.211 \times 0.174}$<br />2. Compute $Q$<br /> $Q = \frac{0.744}{0.036714} = 20.26$<br />3. Apply Gibbs Free Energy Formula<br /> $\Delta G = \Delta G^{\circ} + RT \ln Q$, where $R = 8.314 \, J/(mol \cdot K)$ and $T = 298 \, K$<br />4. Convert Units<br /> $\Delta G^{\circ} = -100.4 \, kJ = -100400 \, J$<br />5. Calculate $\Delta G$<br /> $\Delta G = -100400 + (8.314 \times 298 \times \ln(20.26))$<br />6. Compute $\Delta G$<br /> $\Delta G = -100400 + 8.314 \times 298 \times 3.009$<br /> $\Delta G = -100400 + 7477.9$<br /> $\Delta G = -92822.1 \, J = -92.8221 \, kJ$
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