The first order rate constant for the decomposition of ethyl iodide by the reaction C 2 H 5 I g - C 2 H 5g - HI g at 600 K is 1-60- 10 -5 S -1 Its energy of activation is 209 kJ-mol- Calculate the rate constant of the reaction at 700 K-

logk_2k_1 = E_a2.303 × R (1T_1 1T_2) log k_21.6 ×10^ 5 = 2090002.303 × 8.314 (1600 1700) k_2 = 6.36 ×10^ 3S^ 1 ...

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Standard XII

Chemistry

Rate Constant

The first ord...

Question

The first order rate constant for the decomposition of ethyl iodide by the reaction $C_{2} H_{5} I_{(g)} \to C_{2} H_{5 (g)} + {H I}_{(g)}$ at $600 K$ is ${1.60 \times 10}^{- 5} S^{- 1}$ Its energy of activation is 209 kJ/mol. Calculate the rate constant of the reaction at 700 K.

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Solution

$l o g \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 \times R} (\frac{1}{T_{1}} - \frac{1}{T_{2}})$

$l o g \frac{k_{2}}{1.6 \times 10^{- 5}} = \frac{209000}{2.303 \times 8.314} (\frac{1}{600} - \frac{1}{700})$

$k_{2} = 6.36 \times 10^{- 3} S^{- 1}$

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The first order rate constant for the decomposition of ethyl iodide by the reaction C2H5I(g)→C2H5(g)+HI(g) at 600K is 1.60×10−5S−1 Its energy of activation is 209 kJ/mol. Calculate the rate constant of the reaction at 700 K.

logk2k1=Ea2.303×R(1T1−1T2)logk21.6×10−5=2090002.303×8.314(1600−1700)k2=6.36×10−3 S−1

The first order rate constant for the decomposition of ethyl iodide by the reaction $C_{2} H_{5} I_{(g)} \to C_{2} H_{5 (g)} + {H I}_{(g)}$ at $600 K$ is ${1.60 \times 10}^{- 5} S^{- 1}$ Its energy of activation is 209 kJ/mol. Calculate the rate constant of the reaction at 700 K.

$l o g \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 \times R} (\frac{1}{T_{1}} - \frac{1}{T_{2}})$

$l o g \frac{k_{2}}{1.6 \times 10^{- 5}} = \frac{209000}{2.303 \times 8.314} (\frac{1}{600} - \frac{1}{700})$

$k_{2} = 6.36 \times 10^{- 3} S^{- 1}$