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 60\, sec $

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