For $A B$ , the uncatalysed reaction has $E_{a, fwd} = 85 kJ mo l^{- 1}$ and $E_{a, rev} = 110 kJ mo l^{- 1}$ . A catalyst lowers the forward barrier to $65 kJ mo l^{- 1}$ . For the catalysed reaction

A $Δ H = - 25 kJ mo l^{- 1}$ , $E_{a, rev} = 110 kJ mo l^{- 1}$ and K is unchanged
B $Δ H = - 25 kJ mo l^{- 1}$ , $E_{a, rev} = 90 kJ mo l^{- 1}$ and K increases
C $Δ H = - 25 kJ mo l^{- 1}$ , $E_{a, rev} = 90 kJ mo l^{- 1}$ and K is unchanged
D $Δ H = - 45 kJ mo l^{- 1}$ , $E_{a, rev} = 110 kJ mo l^{- 1}$ and K increases

Answer: C

Uncatalysed

Δ H = 85 - 110 = - 25

kJ/mol. A catalyst does not change the energies of A and B, so

Δ H

and K are unchanged. The reverse barrier drops by the same amount as the forward barrier:

65 + 25 = 90

kJ/mol.