For a given reaction, energy of activation for forward reaction E af is 80 kJ mol -1 and Δ H =-40 kJ mol -1. A catalyst lowers E af to 20 kJ mol -1. The ratio of energy of activation for reverse reaction before and after addition of catalyst is:

Question

For a given reaction, energy of activation for forward reaction E af is 80 kJ mol -1 and Δ H =-40 kJ mol -1. A catalyst lowers E af to 20 kJ mol -1. The ratio of energy of activation for reverse reaction before and after addition of catalyst is: (A) 1.0 (B) 0.5 (C) 1.2 (D) 2.0

Tardigrade Admin · Accepted Answer

Δ H = E f - E b -40=80- E b E b =120 kJ / mol , Catalyst lower the E f to 20 kJ / mol for forward reaction then E f prime=20 kJ / mol We know catalyst decreases the activation energy equal amount in both direction. E b prime=120-60=60 kJ / mol ( E b / E b prime)=(120/60)=2

Q. For a given reaction, energy of activation for forward reaction $E_{a f}$ is $80 k J m o l^{- 1}$ and $Δ H = - 40 k J m o l^{- 1}$ . A catalyst lowers $E_{a f}$ to $20 k J m o l^{- 1}$ . The ratio of energy of activation for reverse reaction before and after addition of catalyst is:

$1.0$

$0.5$

$1.2$

$2.0$

Q. For a given reaction, energy of activation for forward reaction Eaf​ is 80kJmol−1 and ΔH=−40kJmol−1. A catalyst lowers Eaf​ to 20kJmol−1. The ratio of energy of activation for reverse reaction before and after addition of catalyst is:

1.0

0.5

1.2

2.0

ΔH=Ef​−Eb​ −40=80−Eb​ Eb​=120kJ/mol, Catalyst lower the Ef​ to 20kJ/mol for forward reaction then Ef′​=20kJ/mol We know catalyst decreases the activation energy equal amount in both direction. Eb′​=120−60=60kJ/mol Eb′​Eb​​=60120​=2

Q. For a given reaction, energy of activation for forward reaction $E_{a f}$ is $80 k J m o l^{- 1}$ and $Δ H = - 40 k J m o l^{- 1}$ . A catalyst lowers $E_{a f}$ to $20 k J m o l^{- 1}$ . The ratio of energy of activation for reverse reaction before and after addition of catalyst is:

$1.0$

$0.5$

$1.2$

$2.0$