Question

A first-order gas reaction has $k=1.5\times {10}^{-6}{s}^{-1}at{200}^{\circ }C.$ If the reaction is allowed to run for 10 h, what percentage of the initial concentration would have changed in the product? What is the half-life of this reaction? Write in the answer in the form of $100x$?

Answer 1

For a first order reaction,
$k=\frac{2.303}{t}log\frac{A_0}{A_t}$
$=\frac{2.303}{t}log\frac{A_0}{(A_0-x)}$
or $1.5\times {10}^{-6}=\frac{2.303}{3600\times 10s}log\frac{100}{(100-x)}$
or x = 5.2
Thus, the initial concentration changed into product is $5.2\%$.
Half life = $\frac{0.693}{k}=\frac{0.693}{1.5\times {10}^{-6}}$
= 462000 s
= 128.33 hr