Question

For the following reaction:
$A+2B\rightleftharpoons C$
The following data was collected:

[A]	[B]	Rate (M/s)
$0.10$	$0.10$	$3.5\times {10}^{-2}$
$0.20$	$0.10$	$1.4\times {10}^{-1}$
$0.10$	$0.20$	$3.5\times {10}^{-2}$

What is the rate law for this reaction?

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

Answer 1

The correct option is C $Rate=k[A{]}^2$

Let the rate of reaction be given as:

$rate=k\left[A{]}^x\right[B{]}^y$

When a concentration of B kept constant at $0.10$ and the concentration of A changed from $0.10$ to $0.20$, the rate of reaction will be given as

$3.5\times {10}^{-2}=\left[0.10{]}^x\right[0.1{]}^y$-----(1)

$1.4\times {10}^{-1}=\left[0.20{]}^x\right[0.10{]}^y$-----(2)

Dividing equation (1) with (2) we get,

$(2{)}^x=\frac{1.4\times {10}^{-1}}{3.5\times {10}^{-2}}$

$\therefore (2{)}^x=4$

$\therefore x=2$

Similarly when concentration of A kept constant at $0.10$ and concentration of B changes form $0.10$ to $0.20$ we get value of rate remains same.

Thus, the value of rate is independent of the concentration of $B$.

So $Rate=k[A{]}^2$ (not depend on B)