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JEE Main levelSingle correct optionDifficulty 2 of 3ce135

Take $R = 8.314 J mo l^{- 1} K^{- 1}$ and assume $Δ H^{\circ} = + 50.0 kJ mo l^{- 1}$ is constant. By what factor does $K$ change when the temperature is raised from $300 K$ to $320 K$ ?

A $K$ increases by a factor of about 3.5
B $K$ decreases by a factor of about 3.5
C $K$ increases by a factor of about 17.9
D $K$ increases by a factor of about 1.25

Answer: A

ln (K_{2} / K_{1}) = \frac{Δ H ^{\circ}}{R} (\frac{1}{T _{1}} - \frac{1}{T _{2}}) = \frac{50000}{8.314} (\frac{1}{300} - \frac{1}{320}) = 1.253

. So

K_{2} / K_{1} = e^{1.253} = 3.50

. The factor is greater than 1 because the reaction is endothermic.

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