Question

In a chemical reaction A is converted into B. The rates of reaction, starting with initial concentrations of A as $2\times 10^{-3}$ M and $1\times 10^{-3}$ M, are equal to $2.40\times 10^{-4}M{s}^{-1}$ and $0.60\times 10^{-4}M{s}^{-1}$ respectively. The order of reaction with respect to reactant A will be

• A. 0
• B. 1.5
• C. 1
• D. 2

Answer 1

Answer: (D)

$A\to B$
Initial concentration Rate of reaction
$2\times 10^{-3}M2.40\times 10^{-4}M{s}^{-1}$
$1\times 10^{-3}M0.60\times 10^{-4}M{s}^{-1}$
rate of reaction
$r=k{\left[A\right]}^x$
where x = order of reaction
hence
$2.40\times 10^{-4^{}}=k{\left[2\times 10^{-3}\right]}^x......\left(i\right)$
$0.60\times 10^{-4}=k{\left[1\times 10^{-3}\right]}^x......\left(ii\right)$
on dividing eqn. $\left(i\right)$ from eqn. $\left(ii\right)$ we get
$4={\left(2\right)}^x$
$\therefore x=2$
i. e. order of reaction $=2$