What is the concentration (in $\mathrm{mol} \mathrm{L}^{-1}$ ) of the product after $20 \mathrm{~s}$ in the…
What is the concentration (in $\mathrm{mol} \mathrm{L}^{-1}$ ) of the product after $20 \mathrm{~s}$ in the following reaction. Given that $A \longrightarrow 3 B$, rate $=k[\mathrm{~A}]^0$
$6.6 \times 10^{-2}$
$1.32 \times 10^{-1}$
$1.98 \times 10^{-1}$
$2.2 \times 10^{-2}$
Solution
For the zero order reaction, $A \longrightarrow 3 B$ the half-life is given by the formula $t_{1 / 2}=A_0 / 2 k$ where, $A_0$ is the initial concentration and $k$ is the rate constant.
After $15 \mathrm{~s}$ the concentration of $A\left(0.1 \mathrm{~mol} \mathrm{~L}^{-1}\right)$ is reduced to its half value $\left(0.05 \mathrm{~mol} \mathrm{~L}^{-1}\right)$,
so, $t_{1 / 2}=15 \mathrm{~s}$
$
k=0.1 / 2 \times 15=0.0033 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}
$
For zero order reaction after time ' $t$ ' the concentration of reactant $A$ is given by formula
$
[A]=\left[A_0\right]-k t
$
After $t=20 \mathrm{~s}$
$
\begin{aligned}
{[A] } & =0.1 \mathrm{~mol} \mathrm{~L}^{-1}-0.0033 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} \times 20 \mathrm{~s} \\
& =0.033 \mathrm{~mol} \mathrm{~L}^{-1}
\end{aligned}
$
This implies that $(0.1-0.033)=0.067$ moles of reactant $A$ are converted to product $B$.
Since, 1 mole of $A$ gives 3 moles of $B$, so 0.067 moles of reactant $A$ will give $(0.066 \times 3)=0.198$ moles of $B$, i.e. $1.98 \times 10^{-1}$ moles
