Question

For the reaction $2A+B\to C$, the values of initial rate at different reactant concentrations are given in the table below: The rate law for the reaction is:

$\left[A\right]\left(mol{L}^{-1}\right)$	$\left[B\right]\left(mol{L}^{-1}\right)$	$\text{Initial Rate}\left(mol{L}^{-1}{s}^{-1}\right)$
0.05	0.05	0.045
0.10	0.05	0.090
0.20	0.10	0.72

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

Answer 1

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

Let, $r=k\left[A{]}^x\right[B{]}^y$
$0.045=k\left(0.05{)}^x\right(0.05{)}^y$.... (1)
$0.090=k\left(0.10{)}^x\right(0.05{)}^y$..... (2)
$0.72=k\left(0.20{)}^x\right(0.10{)}^y$..... (3)
Dividing (2) by (1),
$2=2^x\Rightarrow x=1$
Dividing (3) by (2),
$8=\left(2^1\right)(2{)}^y\Rightarrow y=2$
So, rate= $k\left[A\right][B{]}^2$