For the reversible reaction:
A(g)+2B(g)⇌C(g)ΔH=−92 kJ mol−1 \text{A}(g) + 2\text{B}(g) \rightleftharpoons \text{C}(g) \quad \Delta H = -92\text{ kJ mol}^{-1} A(g)+2B(g)⇌C(g)ΔH=−92 kJ mol−1the reaction profile diagram below illustrates the pathways for both the uncatalysed and catalysed processes.
The uncatalysed forward reaction has an activation energy, Ea,fE_{\text{a,f}}Ea,f, of +150 kJ mol-1. A catalyst is added which reduces the forward activation energy by 40 kJ mol-1.
Which statement correctly identifies the activation energy of the reverse catalysed reaction (Ea,r,catE_{\text{a,r,cat}}Ea,r,cat) and the effect of the catalyst on the equilibrium constant, KcK_cKc?
Ea,r,cat=+110 kJ mol−1E_{\text{a,r,cat}} = +110\text{ kJ mol}^{-1}Ea,r,cat=+110 kJ mol−1; the value of KcK_cKc increases.
Ea,r,cat=+202 kJ mol−1E_{\text{a,r,cat}} = +202\text{ kJ mol}^{-1}Ea,r,cat=+202 kJ mol−1; the value of KcK_cKc remains unchanged.
Ea,r,cat=+242 kJ mol−1E_{\text{a,r,cat}} = +242\text{ kJ mol}^{-1}Ea,r,cat=+242 kJ mol−1; the value of KcK_cKc remains unchanged.
Ea,r,cat=+202 kJ mol−1E_{\text{a,r,cat}} = +202\text{ kJ mol}^{-1}Ea,r,cat=+202 kJ mol−1; the value of KcK_cKc increases.
