A reaction has forward activation energy 100 kJ/mol and reverse activation energy 60 kJ/mol. Which statements are correct?

AThe forward rate constant is more sensitive to temperature than the reverse rate constant
B $Δ H = + 40 kJ mo l^{- 1}$
CThe catalyst increases the equilibrium constant
DA catalyst that lowers the forward barrier by 30 kJ/mol gives a reverse barrier of 30 kJ/mol

Answer: A, B, D

Δ H = E_{f} - E_{r} = 100 - 60 = + 40

kJ/mol. A catalyst shifts both barriers by the same amount, so the reverse barrier becomes

60 - 30 = 30

kJ/mol, and the equilibrium constant is unchanged. The larger

E_{a}

(forward) gives the larger temperature sensitivity, so statements 1, 2 and 4 are correct.