For the reaction,A rightleftharpoons B,E_a = | vedclass

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(A) The enthalpy change of a reaction is given by the difference between the activation energy of the forward reaction $(E_a)_f$ and the activation en

Vedclass · Accepted Answer

$60 \ kJ \ mol^{-1}$

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(A) The enthalpy change of a reaction is given by the difference between the activation energy of the forward reaction $(E_a)_f$ and the activation energy of the backward reaction $(E_a)_b$: $\Delta H = (E_a)_f - (E_a)_b$.Given that the initial $(E_a)_f = 50 \ kJ \ mol^{-1}$ and the catalyst decreases it by $10 \ kJ \ mol^{-1}$,the new forward activation energy is $(E_a)_f = 50 - 10 = 40 \ kJ \ mol^{-1}$.The enthalpy change $\Delta H$ remains constant at $-20 \ kJ \ mol^{-1}$ because a catalyst does not change the energy of the reactants or products.Substituting the values into the formula: $-20 = 40 - (E_a)_b$.Solving for $(E_a)_b$: $(E_a)_b = 40 + 20 = 60 \ kJ \ mol^{-1}$.

For the reaction,$A \rightleftharpoons B$,$E_a = 50 \ kJ \ mol^{-1}$ and $\Delta H = -20 \ kJ \ mol^{-1}$. When a catalyst is added,$E_a$ decreases by $10 \ kJ \ mol^{-1}$. What is the $E_a$ for the backward reaction in the presence of the catalyst?

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