Question

For a first order reaction, the initial concentration of a reactant is $0.05$ $M$. After $45$ minute it is decreased by $0.015$ $M$. Calculate half-reaction time $\left(t_{1/2}\right)$.

• A. $87.42min$
• B. $25.90min$
• C. $78.72min$
• D. $77.20min$

Answer 1

The correct option is A $87.42min$

For Ist order reaction, the half-life $\left(t_{1/2}\right)$ is:

$t_{1/2}=\frac{0.693}{k}$ ....(i)

$\because k=\frac{2.303}{t}\log \frac{\left[N_0\right]}{N}$

where $N_0=$ Initial concentration
$N=$ Final concentration

$\therefore k=\frac{2.303}{45}\log \frac{0.05M}{0.05-0.015M}$

$k=\frac{2.303}{45}\log \frac{0.05}{0.035}$

$=\frac{2.303}{45}\log \frac{10}{7}$

$=\frac{2.303}{45}[\log 10-\log 7]$

$=\frac{2.303}{45}[1-0.8450]$

$=\frac{2.303}{45}\times 0.154$

$k=0.00792$

Put the value of $k$ in equation (i), we get

$t_{1/2}=\frac{0.693}{0.00792}$

$t_{1/2}=87.5min$

Hence, the correct option is $\text{B}$