The value of log10K for a reaction A⇌ B is___(Given ΔH298K0=−54.07KJ/mol ΔS298K0=10 J/mol−1 R=8.314JK−1mol−1)

ΔG0=ΔH0−TΔS0ΔG0=−2.303RTlog10Klog10K=ΔH−TΔS0−2.303RT=−54.07×103−298×10−2.303×8.314×298log10K=−57050−5705=10

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The value of $\log_{10} K$ for a reaction $A \overset{}{⇌} B$ is___(Given $Δ H_{298 K}^{0} = - 54.07 K J / m o l$ $Δ S_{298 K}^{0} = 10 J / m o l^{- 1}$ $R = 8.314 J K^{- 1} m o l^{- 1}$ )

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Detailed Solution

$Δ G^{0} = Δ H^{0} - T Δ S^{0}$

$Δ G^{0} = - 2.303 R T \log_{10} K$
$\log_{10} K = \frac{Δ H - T Δ S^{0}}{- 2.303 R T}$
$= \frac{- 54.07 \times 10^{3} - 298 \times 10}{- 2.303 \times 8.314 \times 298}$
$\log_{10} K = \frac{- 57050}{- 5705} = 10$

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