Question

The rate constant for the first order decomposition of \(H_2O _2\) is given by the following equation:
\(log\  k = 14.34 - 1.25 \times  10^4\  K/T \)
Calculate \(E_a\) for this reaction and at what temperature will its half-period be 256 minutes?

Answer 1

\(Arrhenius \ equation \  given\  by, \)

\(k = Ae^{\frac {-E_a}{RT}}\)

\(ln\  k = ln \ A - \frac {E_a}{RT}\)

\(ln\  k = log \ A - \frac {E_a}{RT}\)

⇒\(log\  k = log \ A - \frac {E_a}{2.303\ T}\)   .....(i)

The given equation is

\(log\  k = 14.34-1.25×10^4\  K/T\)     ......(ii)

From equation (i) and (ii), we obtain

\(\frac {E_a}{2.303 \ RT} = 1.25×10^4\  K/T\)

⇒ \(Ea=1.25×10^4K×2.303×R\)

= \(1.25 × 10^4K × 2.303 × 8.314 J K^{- 1}mol^ {- 1}\)

= \(239339.3 \ J mol^{-1} (approximately)\)

= \(239.34\  kJ mol^{-1}\)

Also, when \(t_{\frac 12}=256 \ minutes\),

\(k = \frac {0.693}{t_{1/2}}\)

\(k = \frac {0.693}{256}\)

\(k= 2.707×10^{-3} min^{-1}\)

\(k= 4.51 × 10^{-5} s^{-1}\)

It is also given that, \(log\  k= 14.34 - 1.25 × 10^4 K/T\)

⇒\(log(4.51×10^{-5})=14.34-1.25×10^4 K/T\)

\(log \ (0.654-05)=14.34-1.25×10^4 K/T\)

\(\frac {1.25×10^4K}{T} = 18.686\)

⇒\(T =\frac {1.25 \times 10^4 \ K}{18.686}\)

\(T = 668.95 \ K\)

\(T = 669\  K\  (approximately)\)