Question

Ethyl chloride vapor decomposes by the first-order reaction c2h5cl → c2h4 hcl the activation energy is 249 kj/mol and the frequency factor is 1.6 × 1014 s−1. find the temperature at which the rate of the reaction would be twice as fast as when the reaction runs at 730.6 k. enter your answer numerically and in terms of kelvin.

Answer 1

Answer:

The temperature at which the rate of the reaction would be twice as fast as when the reaction runs at 730.6 K is 743.12 K.

Explanation:

According to the Arrhenius equation,

$K=A\times e^{\frac{-Ea}{RT}}$

$\log \frac{K_2}{K_1}=\frac{E_a}{2.303R}\times [\frac{T_2-T_1}{T_1T_2}]$

$K_2$ = rate of reaction at $T_2$

$K_1$ = rate of reaction at $T_1$

$E_a$ = activation energy

R = gas constant  

We have:

$K_1=1.6\times 10^{14} s^{-1}$

$K_2=2\times K_1$

$T_1=730.6 K, T_2=?$

$E_a=249 kJ/mol =249000 J/mol$

$\log \frac{2K_1}{K_1}=\frac{249000 kJ/mol}{2.303\times 8.314 J/mol K}\times [\frac{T_2-730.6 K}{730.6 K\times T_2}]$

$0.3010=\frac{249000 J/mol}{2.303\times 8.314 J/mol K}\times [\frac{T_2-730.6 K}{730.6 K\times T_2}]$

$\frac{0.3010\times 2.303\times 8.314 J/mol K\times 730.6 K\times T_2}{249000 J/mol}=T_2-730.6 K$

$T_2=743.12 K$

The temperature at which the rate of the reaction would be twice as fast as when the reaction runs at 730.6 K is 743.12 K.

Answer 2

Activation Energy Ea = 249 x 10^3 J/mol Frequency Factor = 1.6 Ă— 1014 sâ’1 Rate of reation is twice as fast with temperature change so k1/k2 = 1/2. Temperature T1 = 730.6, T2 =? Gas constant R = 8.314 J/ mole-K We have Arrhenius equation ln (k1 / k2) = (Ea / R) x [(1/T1) - (1/T2)] ln(1/2) = ((249 x 10^3) / 8.314) x [(1/T2) - (1/730.6)] ln(1/2) = 29.95 x 10^3 x [(1/T2) - 0.0013687] -0.69315 / (29.95 x 10^3) = (1/T2) - 0.0013687 (1/T2) - 0.0013687 = - 2.3 x 10^-5 1 / T2 = 0.0013455 T2 = 743.18 k