For the reaction, A �¨ B, ƒ¢H�‹ = +11.00 kJ mol-1 and Keq is 12.50 Calculate i) ƒ¢G�‹ and ii) ƒ¢S�‹ at 25 �‹C. [0 �‹C = 273.15 K; R = 8.314 J K-1 mol-1]
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ƒ¢H�‹
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To calculate ∆G° (standard Gibbs free energy change) and ∆S° (standard entropy change) for the given reaction, we will need to use the relationship:
∆G° = -RT ln(Keq)
∆S° = ∆H° / T - R ln(Keq)
Given:
∆H° = +11.00 kJ mol^(-1)
Keq = 12.50
T = 25 ºC = 25 + 273.15 K = 298.15 K
R = 8.314 J K^(-1) mol^(-1)
i) Calculating ∆G°:
∆G° = -RT ln(Keq)
= -(8.314 J K^(-1) mol^(-1)) * (298.15 K) * ln(12.50)
≈ -8.314 J K^(-1) mol^(-1) * 298.15 K * 2.5250
≈ -6279.69 J mol^(-1)
Since ∆G° is in J, we need to convert it to kJ:
∆G° = -6.2797 kJ mol^(-1) (rounded to four decimal places)
ii) Calculating ∆S°:
∆S° = (∆H° / T) - R ln(Keq)
= ((11.00 kJ mol^(-1)) / (298.15 K)) - (8.314 J K^(-1) mol^(-1)) * ln(12.50)
First, convert ∆H° to J:
∆H° = 11.00 kJ mol^(-1) * 1000 J kJ^(-1)
= 11000 J mol^(-1)
∆S° = (11000 J mol^(-1) / 298.15 K) - (8.314 J K^(-1) mol^(-1)) * ln(12.50)
≈ 36.921 J K^(-1) mol^(-1) - (8.314 J K^(-1) mol^(-1)) * 2.5250
≈ 36.921 J K^(-1) mol^(-1) - 20.9775 J K^(-1) mol^(-1)
≈ 15.9445 J K^(-1) mol^(-1)
Since ∆S° is in J K^(-1) mol^(-1), we need to convert it to kJ K^(-1) mol^(-1):
∆S° = 0.0159 kJ K^(-1) mol^(-1) (rounded to four decimal places)
Therefore, the final answers are:
i) ∆G° = -6.2797 kJ mol^(-1)
ii) ∆S° = 0.0159 kJ K^(-1) mol^(-1)