Question

For the reaction :
$C{H}_{3}COC{H}_3+B{r}_2\stackrel{H^{+}}{\leftarrow }C{H}_{3}COC{H}_{2}Br+H^{+}+B{r}^{-}$

The following data were collected.

$C{H}_{3}COC{H}_3$ $\left[B{r}_2\right]$ $\left[H^{+}\right]$ Rate of disappearance of $B{r}_2$

0.15 0.025 0.025 6.0104
0.15 0.050 0.025 6.0104
0.15 0.025 0.050 1.2103
0.20 0.025 0.100 3.2103
0.20 0.025 0.025 8.0103

The overall order of the reaction is:

• A. 1.5
• B. 4
• C. 5
• D. 2

Answer 1

Answer: (D)

2

Rate of dissappearance of $B{r}^2=k\left[C{H}_{3}COC{H}_3{]}^a\right[B{r}_2{]}^b[H^{+}{]}^c$

$6\cdot 0\times {10}^{-4}=k(0\cdot 15{)}^a(0\cdot 025{)}^b(0\cdot 025{)}^c---(1)$

$6\cdot 0\times {10}^{-4}=k(0\cdot 15{)}^a(0\cdot 050{)}^b(0\cdot 025{)}^c---(2)$

$1\cdot 2\times {10}^{-3}=k(0\cdot 15{)}^a(0\cdot 025{)}^b(0\cdot 050{)}^c---(3)$

$3\cdot 2\times {10}^{-3}=k(0\cdot 20{)}^a(0\cdot 025{)}^b(0\cdot 100{)}^c---(4)$

On dividing, $\left(1\right)÷\left(2\right)$:

$1={\left(\frac{1}{2}\right)}^b$

$b=0$

On dividing, $\left(2\right)÷\left(3\right)$:

$\frac{1}{2}=2^b\cdot {\left(\frac{1}{2}\right)}^c$

$\Rightarrow c=1$

$\frac{1\cdot 2\times {10}^{-3}}{3\cdot 2\times {10}^{-3}}=\left(\frac{0\cdot 15}{0\cdot 20}{)}^a\right(\frac{0\cdot 50}{0\cdot 100})$

$\frac{12}{32}=\left(\frac{3}{4}{)}^a\right(\frac{1}{2})\Rightarrow \frac{3}{4}=(\frac{3}{4}{)}^a$

$\Rightarrow a=1$

Therefore, order of reaction $=a+b+c$

$=2$