Question

The decomposition of A into product has value of k as $4.5\times {10}^3{s}^{-1}$ at ${10}^{\circ }C$ and energy of activation $60kJmo{l}^{-1}$. At what temperature would k be $1.5\times {10}^4{s}^{-1}$?

Answer 1

It is given that the decomposition of $A$ into product has value of $k=4.5\times {10}^3{s}^{-1}$ at ${10}^{o}C$.

$log\left(\frac{K^{\prime }}{k}\right)=\frac{E_a}{2.303R}\left[\frac{T^{\prime }-T}{T{T}^{\prime }}\right]$

$log\left(\frac{1.5\times {10}^4}{4.5\times {10}^3}\right)=\frac{60\times {10}^3}{2.303\times 8.314}\left[\frac{T^{\prime }-283}{283T^{\prime }}\right]$

$0.5228=3132.62\left[\frac{T^{\prime }-283}{283T^{\prime }}\right]$

$T^{\prime }-283=0.0472T^{\prime }$
$T^{\prime }=297.02$ K
$T^{\prime }=297.02K-273=24.02{}^{o}C$