Question

The time required for 10% completion of a first order reaction at 298 K is equal to that required for its 25% completion at 308 K. If the value of A is \(4 \times 10^{10} s^{-1}\) . Calculate \(k\) at 318 K and \(E_a\).

Answer 1

For a first order reaction,
\(t =\frac { 2.303}{k} log\ \frac { a}{a-x}\)
\(At \ 298 \ K\)
\(t =\frac { 2.303}{k} log\ \frac { 100}{90}\)
\(t = \frac {0.1054}{K}\)
\(At\ 308 \ k\)
\(t' =\frac { 2.303}{k'} log\ \frac { 100}{75}\)
\(t' = \frac {2.2877}{k'}\)
According to the question
\(t= t'\)
\(\frac {0.1054}{K} = \frac {2.2877}{k'}\)
\(\frac {k'}{k }= 2.7296\)
From Arrhenius equation, we obtain
\(log \ \frac {k_2}{k_1} = \frac {E_a}{2.303\ R} (\frac {T_2-T_1}{T_1T_2})\)
Therefore, \(log \ (2.7296) = \frac {Ea}{2.303\times8.314} (\frac {308-298}{298\times 308})\)
⇒ \(E_a = \frac {0.6021\times 2.303\times 8.314\times 298\times 308 \times log (2.7296)}{308-298}\)
⇒ \(Ea = 76640.096 \ J mol^{-1}\)
⇒ \(Ea = 76.6\  kJ mol^{-1}\)
\(To\  calculate \ k\  at \ 318 \ K,\)
\(It\  is\  given\  that,\)
\(A = 4\times 10^{10} s^{-1} \ and \ T = 318\  K\)
\(Again,\  from \ Arrhenius \ equation,\  we \ obtain\)
\(log \ k = log\  A- \frac {E_a}{2.303\  RT}\)
\(log\  k = log\ (4\times10^{10}) - \frac {76.64\times 10^3}{2.303\times 8.314×318}\)
\(log \ k = (0.6021+10)-12.5876\)
\(log \ k =-1.9855\)
\(Therefore,\)
\(k = antilog \ (-1.9855)\)
\(k = 1.034\times10^{-2} s^{-1}\)