In a chemical reaction $A$ is converted into $B$. The rates of reaction, starting with initial…
In a chemical reaction $A$ is converted into $B$. The rates of reaction, starting with initial concentrations of $A$ as $2 \times 10^{-3} \mathrm{M}$ and $1 \times 10^{-3}$ $\mathrm{M}$, are equal to $2.40 \times 10^{-4} \mathrm{Ms}^{-1}$ and $0.60 \times 10^{-4} \mathrm{Ms}^{-1}$ respectively. The order of reaction with respect to reactant $A$ will be
0
$1.5$
1
2
Solution
$A \longrightarrow B$
Initial concentration Rate of reaction
$
\begin{array}{ll}
2 \times 10^{-3} \mathrm{M} & 2.40 \times 10^{-4} \mathrm{Ms}^{-1} \\
1 \times 10^{-3} \mathrm{M} & 0.60 \times 10^{-4} \mathrm{Ms}^{-1}
\end{array}
$
rate of reaction
$
r=k[A]^x
$
where $x=$ order of reaction hence
$
\begin{aligned}
& 2.40 \times 10^{-4}=k\left[2 \times 10^{-3}\right]^x \\
& 0.60 \times 10^{-4}=k\left[1 \times 10^{-3}\right]^x
\end{aligned}
$
on dividing eqn.(i) from eqn. (ii) we get $4=(2)^x$
$\therefore \quad x=2$
i.e. order of reaction $=2$