Consider the following reaction at 298 K.32O2( g)⇌O3( g)⋅KP=2.47×10−29.ΔrG⊕ for the reaction is _______ kJ. (Given R =8.314JK−1 mol−1)
Question
Consider the following reaction at 298 K.
for the reaction is _______ kJ. (Given R = 8.314 J K^{-1} mol^{-1})
Solution
The question is asking for the standard Gibbs free energy change (ΔrG⊕) for the given reaction. We can use the relationship between the standard Gibbs free energy change and the equilibrium constant (Kp) at a given temperature. This relationship is given by the equation:
ΔrG⊕ = -RT ln(Kp)
where: R is the universal gas constant = 8.314 J K^-1 mol^-1 T is the temperature in Kelvin = 298 K Kp is the equilibrium constant = 2.47 x 10^-29
Substituting these values into the equation gives:
ΔrG⊕ = - (8.314 J K^-1 mol^-1) * (298 K) * ln(2.47 x 10^-29)
Solving this will give the value of ΔrG⊕ in Joules. To convert this to kJ, divide the result by 1000.
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