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Multiple Choice
Given that the equilibrium constant, Keq, for a reaction at 25°C (298 Kelvin) is 740, what is the standard Gibbs free energy change, ΔG°', for the reaction? Provide your answer to two decimal places.
A
-5.67 kJ/mol
B
25.00 kJ/mol
C
12.34 kJ/mol
D
-17.10 kJ/mol
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검증된 단계별 안내
1
Identify the relationship between the equilibrium constant (Keq) and the standard Gibbs free energy change (ΔG°'). The equation is ΔG°' = -RT ln(Keq), where R is the universal gas constant and T is the temperature in Kelvin.
Determine the value of the universal gas constant, R. For this calculation, use R = 8.314 J/(mol·K).
Convert the temperature from Celsius to Kelvin if necessary. In this case, the temperature is already given as 298 K.
Substitute the known values into the equation: ΔG°' = - (8.314 J/(mol·K)) * (298 K) * ln(740).
Calculate the natural logarithm of the equilibrium constant, ln(740), and then multiply by the other terms to find ΔG°'. Remember to convert the final result from Joules to kilojoules by dividing by 1000.