Question

For a chemical reaction, $2A+2B\to 2C$, the rate controlling step is $A+\frac{1}{2}B\to C$. If the concentration of B is doubled, the rate of reaction will:

• A. remain same
• B. become 4 times
• C. become 1.414 times
• D. become double

Answer 1

Answer: (C)

become 1.414 times

As the rate controlling step is $A+\frac{1}{2}B\to C$.

$Rate=K\left[A\right][B{]}^{\frac{1}{2}}$

$Now,\left[B\right]\to \left[2B\right]$

$(Rate{)}_{new}=K[A\left]\right[2B{]}^{{}^1{/}_2}$$=\sqrt{2}K\left[A\right][B{]}^{{}^1{/}_2}$

The rate become 1.414 times.

Hence, the correct option is $\text{C}$