Question

For a reaction $A+2B\to C+D,$ the following data were obtained
Expt. Initial concentration$\left(moleslitr{e}^{-1}\right)$Initial rate of formation of D $\left(moleslitr{e}^{-1}mi{n}^{-1}\right)$
$\left[A\right]\left[B\right]$
1.$0.10.16.0\times {10}^{-3}$
2.$0.30.27.2\times {10}^{-2}$
3.$0.30.42.88\times {10}^{-1}$
4.$0.40.12.4\times {10}^{-2}$

The correct rate law expression will be:

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

Answer 1

Answer: (B)

(B)$Rate=k\left[A\right]{\left[B\right]}^2$

From experiments (1) and (4), when the concentration of B is kept constant and 0.1 M and the concentration of A is increased four times from 0.1 to 0.4M, the rate of the reaction becomes four times.

$\frac{2.4\times {10}^{-2}}{6.0\times {10}^{-3}}$

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

When the concentration of A is kept constant at 0.3M, and the concentration of B is doubled from 0.2 to 0.4, the rate of the reaction is increased four times.

$\frac{2.88\times {10}^{-1}}{7.2\times {10}^{-2}}=4$

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

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