In an exothermic reaction A to B,the evolved | vedclass

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(B) For an exothermic reaction,the enthalpy change $\Delta_{r} H$ is negative. The relationship between the activation energy of the forward reaction

Vedclass · Accepted Answer

$480$

Vedclass · Answer

(B) For an exothermic reaction,the enthalpy change $\Delta_{r} H$ is negative. The relationship between the activation energy of the forward reaction $(E_{a})_{f}$,the activation energy of the reverse reaction $(E_{a})_{r}$,and the enthalpy change $\Delta_{r} H$ is given by: $\Delta_{r} H = (E_{a})_{f} - (E_{a})_{r}$ Given that the reaction is exothermic,the heat evolved is $280 \ kJ \ mol^{-1}$,so $\Delta_{r} H = -280 \ kJ \ mol^{-1}$. The activation energy of the forward reaction $(E_{a})_{f} = 200 \ kJ \ mol^{-1}$. Substituting these values into the equation: $-280 \ kJ \ mol^{-1} = 200 \ kJ \ mol^{-1} - (E_{a})_{r}$ $(E_{a})_{r} = 200 \ kJ \ mol^{-1} + 280 \ kJ \ mol^{-1} = 480 \ kJ \ mol^{-1}$.

In an exothermic reaction $A \to B$,the evolved heat is $280 \ kJ \ mol^{-1}$ and the activation energy is $200 \ kJ \ mol^{-1}$. The activation energy of the reverse reaction $B \to A$ is $.......... \ kJ \ mol^{-1}$.

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