Ncert Exercises & Solutions

The rate constant for the first order decomposition of H2O2 is given by the following equation:

log k = 14.34 – 1.25 × 10⁴K/T

Calculate Ea for this reaction and at what temperature will its half-period be 256 minutes?

Arrhenius equation is given by

k = A e^{E_{a / R T}}

\Rightarrow \ln k = \ln A - \frac{E_{a}}{R T}

\Rightarrow \ln k = \log A - \frac{E_{a}}{R T}

\Rightarrow \log k = \log A - \frac{E_{a}}{2.303 R T} (i)

The given equation is

\log k = 14.31 - 1.25 \times 10^{4} K / T (i i)

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

\frac{E_{a}}{2.303 R T} = \frac{1.25 \times 10^{4} K}{T}

\Rightarrow E_{a} = 1.25 \times 10^{4} K \times 2.303 \times R

= 1.25 \times 10^{4} K \times 2.303 \times 8.314 J K^{- 1} m o l^{- 1}

= 239339.3 j m o l^{- 1} (a p p r o x i m a t e r l y)

= 239.34 k J m o l^{- 1}

Also, when t

_{1 / 2}

= 256 minutes,

k = \frac{0.693}{t_{\frac{1}{2}}}

= \frac{0.693}{256}

= 2.707 \times 10^{- 3} m i n^{- 1}

= 4.51 \times 10^{- 5} s^{- 1}

It is also given that, log k = 14.34 − 1.25

\times 10^{4} K / T

\Rightarrow \log (4.51 \times 10^{- 5}) = 14.34 - \frac{1.25 \times 10^{4} K}{T}

\Rightarrow \log (0.654 - 05) = 14.34 - \frac{1.25 \times 10^{4} K}{T}

\Rightarrow \frac{1.25 \times 10^{4} K}{T} = 18.686

\Rightarrow T = \frac{1.25 \times 10^{4} K}{18.686}

= 686.95 K

= 669 K (a p p r o x i m a t e l y)