Consider the reaction X leftharpoons Y at 300 K. If Δ H^ θ and K are 28.40 kJ mol ^ -1 and 1.8 × 10^ -7 at the same temperature, then the magnitude of Δ S^ θ for the reaction in J K ^ -1 mol ^ -1 is _______. (Nearest

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Correct answer: 0

The standard Gibbs free energy change is given by:

\Delta G^{\theta} = -RT \ln K = -2.3 RT \log K

First, calculate

\log K

:

\log K = \log(1.8 \times 10^{-7}) = \log(18 \times 10^{-8})

\log K = \log 18 - 8 = \log(2 \times 3^2) - 8

\log K = \log 2 + 2 \log 3 - 8

Substituting the given values:

\log K = 0.30 + 2(0.48) - 8 = 0.30 + 0.96 - 8 = 1.26 - 8 = -6.74

Now, calculate

\Delta G^{\theta}

:

\Delta G^{\theta} = -2.3 \times 8.3 \times 300 \times (-6.74) = 38599.98 \text{ J mol}^{-1}

Using the relation

\Delta G^{\theta} = \Delta H^{\theta} - T \Delta S^{\theta}

:

38599.98 = 28400 - 300 \Delta S^{\theta}

300 \Delta S^{\theta} = 28400 - 38599.98 = -10199.98

\Delta S^{\theta} = -\dfrac{10199.98}{300} \approx -34 \text{ J K}^{-1} \text{ mol}^{-1}

The magnitude of

\Delta S^{\theta}

is

34

.

Answer:

34

Consider the reaction X leftharpoons Y at 300 K. If Δ H^ θ and K...

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