If the activation energy for the forward reaction is 150 text kJ text mol-1 and that of the reverse reaction is 260 text kJ text mol-1, what is the enthalpy change for the reaction?

Question

If the activation energy for the forward reaction is 150 text kJ text mol-1 and that of the reverse reaction is 260 text kJ text mol-1, what is the enthalpy change for the reaction? (A)  410 text kJ text mol-1  (B)  110 kJ mol-1  (C)  -110 kJ mol-1  (D)  - 410 kJ mol-1

Tardigrade Admin · Accepted Answer

Enthalpy change for a reversible reaction, Δ H=Ea(forward)-Ea(backward) Δ H=150-260=-110 text kJ text mol-1

Q. If the activation energy for the forward reaction is $150 k J m o l^{- 1}$ and that of the reverse reaction is $260 k J m o l^{- 1},$ what is the enthalpy change for the reaction?

$410 k J m o l^{- 1}$

$110 k J m o l^{- 1}$

$- 110 k J m o l^{- 1}$

$- 410 k J m o l^{- 1}$

$90 k J m o l^{- 1}$

Enthalpy change for a reversible reaction,
$Δ H = E_{a} (f or w a r d) - E_{a} (ba c k w a r d)$
$Δ H = 150 - 260 = - 110 k J m o l^{- 1}$

Q. If the activation energy for the forward reaction is 150kJmol−1 and that of the reverse reaction is 260kJmol−1, what is the enthalpy change for the reaction?

410kJmol−1

110kJmol−1

−110kJmol−1

−410kJmol−1

90kJmol−1

Enthalpy change for a reversible reaction, ΔH=Ea​(forward)−Ea​(backward) ΔH=150−260=−110kJmol−1