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

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

$=600s$