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 $2.40 \times 10^{-4} \,Ms^{-1}$ and $0.60 \times 10^{-4} \, Ms^{-1}$ 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$\quad\quad$ Rate of reaction
$2 \times 10^{-3} \,M\quad\quad2.40 \times 10^{-4} Ms^{-1}$
$1 \times 10^{-3} \,M\quad\quad0.60 \times 10^{-4} Ms^{-1}$
rate of reaction
$r = k\left[A\right]^{x}$
where x = order of reaction
hence
$2.40 \times 10^{-4^{ }}= k \left[2 \times 10^{-3}\right]^{x}\quad ......\left(i\right)$
$0. 60 \times 10^{-4} = k \left[1 \times 10^{-3}\right]^{x} \quad ......\left(ii\right)$
on dividing eqn. $\left(i\right)$ from eqn. $\left(ii\right)$ we get
$4 = \left(2\right)^{x}$
$\therefore \quad x = 2$
i. e. order of reaction $= 2$