Answer to Question #139312 in Physical Chemistry for Malika

Question

As we know the thermal decomposition of ethyl bromide is complex, the overall
rate is first-order, and the lab constant is given by the expression k = (3.8 x
10^14 s-1)e-229000/RT, where the activation energy is in J mol-1. Calculate the
temperature at which (a) ethyl bromide decomposes at the rate of 1% per
second and (b) the decomposition is 70% complete in 1 hour.

Answer 1

For the first order reaction, $[X]=[X]_ie^{-kt}$

$(a)$ $[X]=\frac{99}{100}[X]_i$

$ln(\frac{99}{100})=-kt$ at $t=1 sec$

$-0.01005=(-3.8\times10^{14})e^{-\frac{229000}{RT}}$

Taking log both sides;

$-4.600=1.335+32.236-\frac{229000}{RT}$

$\frac{229000}{8.314\times T}=38.171$

Hence, $T=721.588K$

$(b)$ $ln(0.3)=-(3.8\times10^{14})e^{-\frac{229000}{RT}}\times3600$

$-1.2039=-1368\times10^{15}\times e^{-\frac{229000}{RT}}$

$8.8\times10^{-19}=e^{-\frac{229000}{RT}}$

$16.825=\frac{229000}{8.314\times T}$

Hence, $T=1637.05K$