Variation of equilibrium constant ‘K’ with temperature ‘T’ is shown as below
Given ‘OP’ = 10 and $\tan θ = 0.5; R = 8.314 J / K^{- 1} m o l^{- 1}$
The equilibrium constant (K) at 298K (T) is

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$9.96 \times 10^{2}$

$9.96 \times 10^{9}$

$9.96 \times 10^{6}$

$9.96 \times 10^{8}$

answer is A.

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Detailed Solution

$\log k = \log A - \frac{Δ H^{0}}{2.303 R T}$

$∴$ Slope $= \frac{- Δ H^{0}}{2.303 R} = - 0.5$

$∴ Δ H^{0} = 2.303 \times 8.314 \times 0.5 = 9.574 J / m o l e$

$∴ \log_{10} K = 10 - \frac{9.574}{2.303 \times 8.314 \times 298} = 9.998$

$∴ K = 9.96 \times 10^{9}$

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