Question

A first order reaction has $k=1.5\times {10}^{-6}{s}^{-1}$ at ${200}^{o}C$. If the reaction is allowed to run for $10hr$, what percentage of the initial concentration would have changed in the product? What is the half life (in hr) of this reaction?

Answer 1

For a first order reaction,

$k=\frac{2.303}{t}\log \frac{a}{a-x}$

Given: $t=10\times 60\times 60s$
Let, initial concentration $\left(a\right)=1$

$\therefore k=\frac{2.303}{10\times 60\times 60}\log \frac{1}{(1-x)}$

$1.5\times {10}^{-6}=\frac{2.303}{10\times 60\times 60}\log \frac{1}{(1-x)}$

$or\log \frac{1}{(1-x)}=\frac{1.5\times {10}^{-6}\times 10\times 60\times 60}{2.303}$

$\log \frac{1}{(1-x)}=0.0234$

$or\frac{1}{(1-x)}=1.055$

$1.055–1.055x=1$

$\therefore x=\frac{1.055-1}{1.055}=0.052$

Thus, $5.2\%(0.052\times 100=5.2)$ of the initial concentration of reactants has changed into products.

Since, $t_{1/2}=\frac{0.693}{k}=\frac{0.693}{1.5\times {10}^{-6}}$

$\therefore t_{1/2}=462000sor128.33hr$