Question

A first order reaction has $k=1.5\times {10}^{-6}$ per second at ${200}^{0}C$. If the reaction is allowed to run for 10 hrs, what percentage of the initial concentration would have changed into the product?

• A. 5.213%
• B. 10.21%
• C. 20.3%
• D. 15.2%

Answer 1

The correct option is A 5.213%

Given,

$K=1.5\times {10}^{-6}{s}^{-1}$

We know for a first order reaction,

$t=\frac{2.303}{K}log\left(\frac{N_0}{N}\right)$

So on substituting given values,

$\frac{N_0}{N}=1.0555$

$⟹N=0.9475N_0$

Now product formed$=0.9475N_0-N_0=0.0521N_0=$5.213%