Question

The rate constant of the reaction $A\to 2B$ is $1.0\times {!0}^{-3}mol$ ${lit}^{-1}$ ${min}^{-1}$, if the initial concentration of $A$ is $1.0mole$ ${lit}^{-1}$, what would be the concentration of $B$ after $100$ minutes?

• A. $0.1mol$ ${lit}^{-1}$
• B. $0.2mol$ ${lit}^{-1}$
• C. $0.9mol$ ${lit}^{-1}$
• D. $1.8mol$ ${lit}^{-1}$

Answer 1

The correct option is B $0.2mol$ ${lit}^{-1}$

The unit of rate constant is given by $mo{l}^{1-n}li{t}^{n-1}mi{n}^{-1}$. From the given units, $1-n=1\Rightarrow n=0$ i.e. zero order.

$\therefore K_t=a_o-a_t$

Given $a_o=1,K=1\times {10}^{-3}molli{t}^{-1}mi{n}^{-1},t=100$ min

$\therefore a_t=a_o-K_t=1-100\times {10}^{-3}=0.9$

Concentration of A= $-0.9$

$\therefore$ Concentration of B= $2(1-0.9)=0.2molli{t}^{-1}$ .