Question

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

• A. $0.052\%$
• B. $52\%$
• C. $5.20\%$
• D. $2.6\%$

Answer 1

Answer: (C)

For a first order reaction

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

$Given:t=10\times 60\times 60sec$

Let initial concentration $(a)$ be 1 , Then, $k=\frac{2.303}{10\times 60\times 60}\log \frac{1}{(1-x)}$

$\Rightarrow \log \frac{1}{(1-x)}=0.0234$ or $\frac{1}{(1-x)}=1.055$

or $x=\frac{1.055-1}{1.055}=0.052$ i.e., $0.052\times 100=5.2\%$