The diagram below shows a general enthalpy profile for a reversible exothermic reaction:
Consider the following specific chemical equilibrium:
P(g)+2Q(g)⇌2R(g)ΔH=−85 kJ mol−1 \text{P}(g) + 2\text{Q}(g) \rightleftharpoons 2\text{R}(g) \quad \Delta H = -85\text{ kJ mol}^{-1} P(g)+2Q(g)⇌2R(g)ΔH=−85 kJ mol−1The activation energy for the uncatalysed forward reaction is +145 kJ mol-1. A homogeneous catalyst is added to the system, which reduces the activation energy of the forward reaction by 35 kJ mol-1.
Which statement is correct?
The catalyst increases the rate of the forward reaction more than the reverse reaction, temporarily increasing the yield of R\text{R}R.
The activation energy of the catalysed reverse reaction is +110 kJ mol−1+110\text{ kJ mol}^{-1}+110 kJ mol−1.
The catalyst decreases the activation energy of the reverse reaction by 35 kJ mol−135\text{ kJ mol}^{-1}35 kJ mol−1, making the new activation energy +195 kJ mol−1+195\text{ kJ mol}^{-1}+195 kJ mol−1.
Since the forward reaction is exothermic, adding the catalyst causes the value of the equilibrium constant, KcK_cKc, to decrease.
