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# The decomposition of hydrogen bromide into hydrogen and iodine is second order reaction.The reaction rate constant is $K=0.08Lmol^{-1}s^{-1}$.How long it take an initial concentration of 0.04M to decrease to half this concentration?

$\begin{array}{1 1}(a)\;300sec&(b)\;315sec\\(c)\;212.5sec&(d)\;312.5sec\end{array}$

$2HBr\rightarrow H_2+Br_2$
At $t=0$ $2HBr\Rightarrow a$
At $t=t$ $2HBr\Rightarrow a-x,H_2\Rightarrow x/2,Br_2\Rightarrow x/2$
Rate $=\large\frac{dx}{2dt}$$=K(a-x)^2 \large\frac{dx}{(a-x)^2}$$=2Kdt$
$\large\frac{dx}{(a-x)^2}$$=K'dt 2K=K' K'=rate constant=0.08Lmol^{-1}s^{-1} -\large\frac{1}{a-x}\bigg]_0^{a/2}$$=K't\big]_0^{t_{1/2}}$
$-\big[\large\frac{1}{a}-\frac{1}{a/2}\big]$$=k't_{1/2}$
$t_{1/2}=\large\frac{1}{k'a}$
$t_{1/2}=\large\frac{1}{0.08\times 0.04}$
$t_{1/2}=312.5sec$
Hence (d) is the correct answer.