Question

A first order reaction is 50% completed in $1.26\times {10}^{14}$. How much time would it take for 100% completion?

• A. $1.26\times {10}^{15}s$
• B. Infinite
• C. $2.52\times {10}^{14}s$
• D. $2.52\times {10}^{28}s$

Answer 1

The correct option is B Infinite

Given half life of first order

$t_{50}=1.26\times {10}^{14}s$

The relation between half life of first order and rate constant is given as:

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

Where,

$t_{\frac{1}{2}}$= half life of first order reaction

k= Rate constant of first order reaction

Thus, $k=\frac{0.693}{1.26\times {10}^{14}}=0.55\times {10}^{14}{s}^{-1}$

$t=\frac{1}{k}in\frac{\left[A_0\right]}{\left[A_t\right]}$

Where,
t= time
k=Rate constant
$A_0$=Initial concentration of reactant
$\left[A_t\right]$= Reactant concentration at time‘t’
Therefore, at 100% completion of reaction:

$A_0=100$

$\left[At\right]=0$

And,

$t=\frac{2.303}{0.55\times {10}^{-14}}\times log\frac{100}{0}$

$t=\infty$

Thus, time taken for 100% completion of a reaction is infinite.

Hence, the option (D) is correct.