Ncert Exercises & Solutions

The rate constant for the decomposition of hydrocarbons is 2.418 × 10^–5s^–1 at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponential factor.

k = 2.418 \times 10^{- s} s^{- 1}

T = 546 K

E_{a} = 179.9 k l m o l^{- 1} = 179.9 \times 10^{3} j m o l^{- 1}

According to the Arrhenius equation,

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

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

\Rightarrow \log A = \log k + \frac{E_{a}}{2.303 R T}

= \log (2.418 \times 10^{- 5} s^{- 1}) + \frac{179.9 \times 10^{3} J m o l^{- 1}}{2.303 \times 8.314 J K^{- 1} \times 546 K}

= (0.3835 - 5) + 17.2082

= 12.5917

Therefore, A = antilog (12.5917)

3.9 \times 10^{12} s^{- 1}

(approximately)

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