Solveeit Logo

Question

Question: For a reactions \(A + B \rightarrow\)product, it was found that rate of reaction increases four time...

For a reactions A+BA + B \rightarrowproduct, it was found that rate of reaction increases four times if concentration of ‘A’ is doubled, but the rate of reaction remains unaffected. If concentration of ‘B’ is doubled. Hence, the rate law for the reaction is.

A

rate=k[A][B]\text{rate} = k\lbrack A\rbrack\lbrack B\rbrack

B

rate=k[A]2\text{rate} = k\lbrack A\rbrack^{2}

C

rate=k[A]2[B]1\text{rate} = k\lbrack A\rbrack^{2}\lbrack B\rbrack^{1}

D

rate=k[A]2[B]2\text{rate} = k\lbrack A\rbrack^{2}\lbrack B\rbrack^{2}

Answer

rate=k[A]2\text{rate} = k\lbrack A\rbrack^{2}

Explanation

Solution

Let the rate of reaction depends on xth power of [A].

Then r1=k[A]xr_{1} = k\lbrack A\rbrack^{x} and r2=k[2A]xr_{2} = k\lbrack 2A\rbrack^{x}

r1r2=[A]x[2A]x=14=(12)2\therefore\frac{r_{1}}{r_{2}} = \frac{\lbrack A\rbrack^{x}}{\lbrack 2A\rbrack^{x}} = \frac{1}{4} = \left( \frac{1}{2} \right)^{2} (r2=4r1)(\because r_{2} = 4r_{1})

x=2\therefore x = 2. As the reaction rate does not depend upon the concentration of B. Hence, the correct rate law will be rate=K[A]2[B]o\text{rate} = K\lbrack A\rbrack^{2}\lbrack B\rbrack^{o} or =K[A]2= K\lbrack A\rbrack^{2}