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Enthalpy

The activatio...

Question

The activation energy for forward and backward reactions are $50 k J / m o l$ and $40 k J / m o l$ respectively. If $K_{1}$ and $K_{2}$ are the equilibrium constants of reaction at temperature $T_{1}$ and $T_{2}$ respectively and $T_{2} > T_{1}$ then:

A

K_{1} < K_{2}

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B

K_{1} = K_{2}

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C

K_{1} > K_{2}

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D

K_{2} = K_{1}^{2}

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Solution

The correct option is A $K_{1} < K_{2}$
The activation energy for forward and backward reactions are $50 k J / m o l$ and $40 k J / m o l$ respectively. If $K_{1}$ and $K_{2}$ are the equilibrium constants of reaction at temperature $T_{1}$ and $T_{2}$ respectively and $T_{2} > T_{1}$ then $K_{1} < K_{2}$ .
The equilibrium constant at higher temperatures is higher than the equilibrium constant at lower temperatures.
$Δ H = E_{f} - E_{b} = 50 - 40 = 10 k J / m o l$ . The positive value of the enthalpy change of reaction suggests endothermic nature.
In endothermic reactions, equilibrium constant increases with an increase in temperature.

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