Question

A first order reaction has $k=1.5\times {10}^{-6}$ per second at ${200}^o$C. If the reaction is allowed to run for $10$ hrs, what percentage of the initial concentration would have changed into the product? What is the half life reaction.

Answer 1

For a first reaction, we have $k=\frac{2.303}{t}\log \frac{{\left[A\right]}_o}{{\left[A\right]}_t}$$

Let ${\left[A\right]}_o$ be 'a'. If change in original conc. of [A] is x, then ${\left[A\right]}_t$ can be written as $a-x.$

$\therefore k=\frac{2.303}{t}\log \frac{a}{a-x}⟶\left(1\right)$

Let initial concentration be $a=1.$

substituting values in 1, we get

$1.5\times {10}^{-6}=\frac{2.303}{10\times 3600}\log \frac{1}{1-x}\frac{1.5\times {10}^{-6}\times 10\times 3600}{2.303}=\log \frac{1}{1-x}=0.0234\frac{1}{1-x}=antilog\left(0.234\right)=1.0554x=\frac{0.0554}{1.0554}=5.25$

Half life of the reaction

$t_{1/2}=\frac{0.693}{k}$

substituting k, we get

$t_{1/2}=\frac{0.693}{1.5\times {10}^{-6}}=4.62\times {10}^{5}seconds$