Answer:
a) at 25°C the rate of reaction increases by a factor of 1,027*10^7
b) at 25°C the rate of reaction increases by a factor of 1,777*10^5
Explanation:
using the Arrhenius equation
k= ko*e^(-Ea/RT)
where
k= reaction rate
ko= collision factor
Ea= activation energy
R= ideal gas constant= 8.314 J/mol*K
T= absolute temperature
for the uncatalysed reaction
k1= ko*e^(-Ea1/RT)
for the catalysed reaction
k2= ko*e^(-Ea2/RT)
dividing both equations
k2/k1= e^(-(Ea2-Ea1)/RT)
a) at 25°C
k2/k1 = e^(-(55kJ/mol-95kJ/mol)/(8.314J/mol*K*298K)* (1000J/kJ ) ) = 1,027*10^7
therefore at 25°C , k2/k1 = 1,027*10^6
b) at 125°C
k2/k1 = e^(-(55kJ/mol-95kJ/mol)/(8.314J/mol*K*298K)* (1000J/kJ ) ) = 1,777*10^5
therefore at 125°C , k2/k1 = 1,777*10^5
Note:
when the catalysts is incorporated, the catalysed reaction and the uncatalysed one run in parallel and therefore the real reaction rate is
k real = k1 + k2 = k2 (1+k1/k2)
since k2>>k1 → 1+k1/k2 ≈ 1 and thus k real ≈ k2
