Question

For a chemical reaction the energy of activation is $85$ $kJ$ ${mol}^{-1}$. If the frequency factor is $4.0\times {10}^9$ $L$ ${mol}^{-1}{s}^{-1}$, what is the rate constant at $400$ $K$?

Answer 1

$E_a=850Kj/mol$

$A=4\times {10}^{9}L{mol}^{-1}{s}^{-1}$

$logK=logA-E_a/RT\times 2.303$

$i.e.$

$logK=logA-\frac{E_a}{2.303\times R\times T}$

$logK=log(4\times {10}^9)-\frac{85\times 1000}{2.303\times 8.314\times 400}$

$logK=9.60-11.09$

$K=0.0323{mol}^{-1}{ls}^{-1}$