Question

In the reaction A + 2B $⟶$ 2C + D. if the concentration of A is increased four times and B is decreased to half of its initial concentration then the rate becomes:

• A. twice
• B. half
• C. unchanged
• D. one fourth of the rate

Answer 1

Answer: (C)

unchanged

The given reaction is $A+2B⟶2C+O$

Rate law is given by:-

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

Now, if the concentration of $A$ is increased $4$ times & concentration of $B$ is increased $1/2$ of the initial concentration. Then,

$(Rate{)}_{New}=[4A\left]\right[B/2{]}^2$

$=4\left[A\right]\frac{[B{]}^2}{4}$

$\Rightarrow (Rate{)}_{New}=[A\left]\right[B{]}^2$ $-\left(ii\right)$

$\left(i\right)$ & $\left(ii\right)\Rightarrow$ Rate is unchanged