Question

For the reaction, $2A+B\to Products,$ the following initial rates were obtained at various initial concentrations:

$\left[A\right]\left[B\right]Rate\left(mol{L}^{-1}se{c}^{-1}\right)$
$0.1M0.2M0.46$
$0.2M0.2M1.84$
$0.2M0.1M0.92$

The rate law for the reaction is:

• A. (A)$Rate=k{\left[A\right]}^2{\left[B\right]}^0$
• B. (B)$Rate=k\left[A\right]\left[B\right]$
• C. (C)$Rate=k{\left[A\right]}^2\left[B\right]$
• D. (D)$Rate=k\left[A\right]{\left[B\right]}^2$

Answer 1

The correct option is C (C)$Rate=k{\left[A\right]}^2\left[B\right]$

From experiments (1) and (2), when the concentration of $B$ is kept constant at $0.2$ and the concentration of $A$ is doubled from $0.1$ to $0.2$, the rate of the reaction becomes four times.

$\frac{1.84}{0.46}=4$

Hence, the order of the reaction w.r.t $A$ is $2$.

From experiments (3) and (2), when the concentration of A is kept constant at $0.2$, and the concentration of B is doubled from $0.1$ to $0.2$, the rate of the reaction is doubled.

$\frac{1.84}{0.92}=2$

Hence, the order of the reaction w.r.t $B$ is $1$.

Hence, the rate law expression is $rate=k\left[A\right]{[}^2\left[B\right]$

Option C is correct.