Question

The rate constant for the reaction below was determined to be 3.241×10-5 s–1 at 800 K. The activation energy of the reaction is 215 kJ/mol. What would be the value of the rate constant at 9.10×102 K? N2O(g) --> N2(g) + O2(g)

I'm having trouble calculating the rate constant with the arrhenius equation that deals with two temps, could you show me the step by step how to do this?

Answer #1

Ans :- **1.6 x 10 ^{-3} s^{-1}**

**Explanation :-**

Given,

rate constant at 800 K = k_{1} = 3.241 x 10^{-5}
s^{-1}

Temperature = T_{1} = 800 K

rate constant at 9.10 x 10^{2} K = k_{2} =?

Temperature = T_{2} = 9.10 x 10^{2} K

Activation energy = E_{a} = 215 KJ = 215000 J/mol

From the Arrhenius equation

ln (k_{2}/k_{1}) = (E_{a}/R) .
(1/T_{1} - 1/T_{2})

ln (k_{2}/k_{1}) = 215000 J/mol / 8.314 J
K^{-1} mol^{-1} .(9.10 x 10^{2} - 800 K /
(800 K)(9.10 x 10^{2})

ln (k_{2}/k_{1}) = 3.9074

k_{2} = k_{1} x exp ( 3.9075)

k_{2} = 3.241 x 10^{-5} x 49.7694

**k _{2} = 1.61 x 10^{-3}
s^{-1}**

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is typically written as
k=Ae−Ea/RT
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