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Q.

If the activation energy for the forward reaction is $150 kJ {mol}^{- 1}$ and that of the reverse reaction is $260 kJ {mol}^{- 1}$ . What is the enthalpy change for the reaction

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a

$410 kJ {mol}^{- 1}$

b

$- 110 kJ {mol}^{- 1}$

c

$110 kJ {mol}^{- 1}$

d

$- 410 kJ {mol}^{- 1}$

answer is B.

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Detailed Solution

Enthalpy change for a reversible reaction,

$\begin{aligned} ΔH = E_{a} (forward) - E_{a} (backward) \\ ΔH = 150 - 260 = - 110 kJ {mol}^{- 1} \end{aligned}$

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