Question

After how many seconds will the concentration of the reactant in a first order reaction be halved if the rate constant is $ 1.155 \times 10^{-3}\, s^{-1} $ ?

• A. $ 600 $
• B. $ 100 $
• C. $ 60 $
• D. $ 10 $

Answer 1

Answer: (A)

Rate constant $k=1.155 \times 10^{-3} s ^{-1}$

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

$\because a=a,(a-x)=\frac{a}{2}$

$t_{1 / 2} =\frac{2.303}{k} \log \frac{a}{a / 2}$

$=\frac{2.303}{1.155 \times 10^{-3}} \log 2$

$=\frac{2.303}{1.155 \times 10^{-3}} \times 0.3010$

$=\frac{0.693 \times 10^{3}}{1.155}$

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

$= \frac{0.693}{1.155 \times 10^{-3}}$

$=600 s$