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Answer :
B. 14.31 kcal/mol is the correct option.
The enthalpy change (∆H) for the reaction can be calculated using the equation:
∆G = -RT ln(Kp)
where:
∆G is the Gibbs free energy change
R is the gas constant (given as 2 cal/mol-K)
T is the temperature in Kelvin
Kp is the equilibrium constant.
Given that Kp = 1 at 27°C and Kp = 4 at 47°C, we can find the change in ∆G using these values.
Using the equation ∆G = ∆H - T∆S, and assuming ∆S is constant, we can find ∆H by rearranging terms.
At 27°C:
∆G₁ = -2 * 27 * ln(1) = 0
At 47°C:
∆G₂ = -2 * 47 * ln(4)
Thus, ∆H = ∆G₂ - ∆G₁ = -2 * 47 * ln(4) - 0
Calculating this gives us the value for ∆H, which is approximately 14.31 kcal/mol. This value represents the enthalpy change for the given reaction. So, the correct option is B. 14.31 kcal/mol.
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