Question

. The first order rate constant for the decomposition of ethyl iodide by the reactionC2H5I(g) ‚Üí C2H4 (g) + (g) at 600K is 1.60 √ó 105 s1. Its energy of activation is 209 kJ/mol.Calculate the rate constant of the reaction at 700K.

Answer 1

Answer: The rate constant of the reaction at 700 is $1.56\times 10^{-1} sec^{-1}$

Explanation:

$\log \frac{k_2}{k_1}=\frac{E_a}{2.303 R}\times [\frac{1}{T_1}-\frac{1}{T_2}]$

$k_1=1.60\times 10^{-5} sec^{-1}$

$k_2=?$

$T_1=600 K,T_2=700 K, R=8.314 J/(K mol)$

$E_a=209 KJ/mol=209000 J/mol$ (1000 J = 1 kJ)

On substituting the above values is formula:'

we get:

$k_2=1.56\times 10^{-1} sec^{-1}$

The rate constant of the reaction at 700 is $1.56\times 10^{-1} sec^{-1}$

Answer 2

logk1k2=2.303×REa(T11−T21)

log1.6×10−5k2=2.303×8.314209000(6001−7001)

k2=6.36×10−3 S−1