Question

$[C{H}_3-\stackrel{O}{\stackrel{||}{C}}-C{H}_3],M$	$\left[I_2\right],M$	$\left[H^{+}\right],M$	Relative rate
0.010	0.010	0.010	1
0.020	0.010	0.010	2
0.020	0.020	0.010	3
0.020	0.010	0.020	4

Propanone reacts with iodine in acid medium according to the following equation:

$C{H}_3-\stackrel{O}{\stackrel{||}{C}}-C{H}_3+I_2\stackrel{H^{+}}{\to }C{H}_3-\stackrel{O}{\stackrel{||}{C}}-C{H}_{3}I+HI$

The above data were obtained when the reaction was studied.
What is the rate equation for the reaction?

• A. Rate $=k[C{H}_3-\stackrel{O}{\stackrel{||}{C}}-C{H}_3]\left[I_2\right]$
• B. Rate $=k[C{H}_3-\stackrel{O}{\stackrel{||}{C}}-C{H}_3{]}^2$
• C. Rate $=k[C{H}_3-\stackrel{O}{\stackrel{||}{C}}-C{H}_3]\left[I_2\right]\left[H^{+}\right]$
• D. Rate $=k[C{H}_3-\stackrel{O}{\stackrel{||}{C}}-C{H}_3]\left[H^{+}\right]$

Answer 1

The correct option is D Rate $=k[C{H}_3-\stackrel{O}{\stackrel{||}{C}}-C{H}_3]\left[H^{+}\right]$

Let rate $=k[C{H}_3-C(=O)-C{H}_3{]}^x[I_2{]}^y[H^{+}{]}^z$

$1=k\left(0.01{)}^x\right(0.01{)}^y(0.01{)}^z$-(1)

$2=k\left(0.02{)}^x\right(0.01{)}^y(0.01{)}^z$-(2)

$3=k\left(0.02{)}^x\right(0.02{)}^y(0.01{)}^z$-(3)

$4=k\left(0.02{)}^x\right(0.01{)}^y(0.02{)}^z$-(4)

From (1) and (2), $\frac{1}{2}=(\frac{1}{2}{)}^x=>x=1$

From (2) and (4), $\frac{2}{4}=(\frac{2}{4}{)}^z=>z=1$

From (2) and (3), $\frac{2}{3}=(\frac{1}{2}{)}^y$

Therefore, y is equal to 0

Rate=$k[C{H}_3-C(=0)-C{H}_3]\left[H^{+}\right]$