Question

For reaction,
$aA+bB\to \text{Product}$
$\text{Rate of reaction}=K{\left[A\right]}^a{\left[B\right]}^b$.
If concentration of 'A' is doubled, rate is increased to four times. If concentration of B is made four times, rate is doubled. What is relation between rate of disappearance of A and that of B?

• A. $-\left\{\frac{d\left[A\right]}{dt}\right\}=-\left\{\frac{d\left[B\right]}{dt}\right\}$
• B. $-\left\{\frac{d\left[A\right]}{dt}\right\}=-\left\{\frac{4d\left[B\right]}{dt}\right\}$
• C. $-\left\{\frac{4d\left[A\right]}{dt}\right\}=-\left\{\frac{d\left[B\right]}{dt}\right\}$
  
• D. None of these

Answer 1

The correct option is B $-\left\{\frac{d\left[A\right]}{dt}\right\}=-\left\{\frac{4d\left[B\right]}{dt}\right\}$

Initial rate method:
$R=K\left[A{]}^p\right[B{]}^q[C{]}^r$
p, q, and r are order of reaction with respect to A, B and C respectively.

When two different intial concentration of A,
$[A_o{]}_1$ and $[A_o{]}_2$ are taken

$r_1=K^{\prime }[A_o{]}_1^p$
$r_2=K^{\prime }[A_o{]}_2^p$

$\frac{r_1}{r_2}=(\frac{[A_o{]}_1}{[A_o{]}_2}{)}^p$

From this relation 'p' can be calculated.

For given reaction
$aA+bB\to Product$

$Rate=K{\left[A\right]}^a{\left[B\right]}^b$
Let us consider, rate of reaction is 'R'
On doubling concentration of A,
Concentration of A is 2A
Rate of reaction is 4R

By initial rate method,
$\frac{R}{4R}=(\frac{A}{2A}{)}^a$
$\frac{1}{4}=(\frac{1}{2}{)}^a$
Thus, $a=2$

Similarly when 'B' is made to four times, the rate of reaction is doubled.
$\frac{R}{2R}=(\frac{B}{4B}{)}^b$
$\frac{1}{2}=(\frac{1}{4}{)}^b$
Thus, $b=\frac{1}{2}$

Hence, the rate of reaction (ROR) is
$Rate=K{\left[A\right]}^2{\left[B\right]}^{\frac{1}{2}}$
$2A+\frac{1}{2}B\to Product$

$\because ROR=\frac{-1}{a}\frac{d\left[A\right]}{dt}=-\frac{1}{b}\frac{d\left[B\right]}{dt}$

$\therefore ROR=\frac{-1}{2}\frac{d\left[A\right]}{dt}=-\frac{1}{\frac{1}{2}}\frac{d\left[B\right]}{dt}$

$\frac{-d\left[A\right]}{dt}=-4\frac{d\left[B\right]}{dt}$