For the reaction PCl5(g) → PCl3(g) + Cl2(g) at298 K, Keq = 1.87 × 10-7, what is ΔrG ° and isthe reaction spontaneous?
Question
For the reaction PCl5(g) → PCl3(g) + Cl2(g) at298 K, Keq = 1.87 × 10-7, what is ΔrG ° and isthe reaction spontaneous?
Solution
To find the ΔrG° (standard Gibbs free energy change), we can use the equation:
ΔrG° = -RT ln Keq
Where: R is the universal gas constant, which is 8.314 J/(mol·K) T is the temperature in Kelvin, which is given as 298 K Keq is the equilibrium constant, which is given as 1.87 × 10^-7
Substituting the given values into the equation, we get:
ΔrG° = - (8.314 J/(mol·K)) * (298 K) * ln (1.87 × 10^-7) ΔrG° = - (8.314 J/(mol·K)) * (298 K) * (-16.19) ΔrG° = 39936.8 J/mol ΔrG° = 39.94 kJ/mol
The sign of ΔrG° determines whether the reaction is spontaneous. If ΔrG° is negative, the reaction is spontaneous. If ΔrG° is positive, the reaction is non-spontaneous. In this case, ΔrG° is positive, so the reaction is non-spontaneous.
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