Question

The gas phase decomposition of acetic acid at 1189 K proceeds by way of two parallel reactions.
 $C{H_3}COOH \to C{H_4} + C{O_2}\,{k_1} = 3.8{\sec ^{ - 1}}$
 $C{H_3}COOH \to C{H_2} = C = O + {H_2}O\,{k_2} = 0.2{\sec ^{ - 1}}$
What is the maximum percentage yield of the ketene $C{H_2}CO$ obtainable at this temperature?
A. \[5\% \]
B. \[95\% \]
C. \[15\% \]
D. \[10\% \]

Answer 1

Hint: The two parallel reactions are given. The ratio of concentration of this reaction is following:
 $\dfrac{{[B]}}{{[C]}} = \dfrac{{k1}}{{k2}}\,or\,\dfrac{{[C]}}{{[B]}} = \dfrac{{k2}}{{k1}}$

Complete step by step solution:
We have
 ${k_1} = 3.8{\sec ^{ - 1}}$
 ${k_2} = 0.2{\sec ^{ - 1}}$
This equation is used for rate of change of Acetic acid
 \[{[C{H_2}CO]_\infty } = \dfrac{{{k_1}}}{{{k_1} + {k_2}}} \times {[C{H_3}COOH]_0}\]
$\Rightarrow$ $\dfrac{{{{[C{H_2}CO]}_\infty }}}{{{{[C{H_3}COOH]}_0}}} = \dfrac{{{k_1}}}{{{k_1} + {k_2}}} \times 100\% $
$\Rightarrow$ $\dfrac{{[C{H_2} = C = 0]}}{{[C{H_4}] \times 100 = \dfrac{{{k_2}}}{{{k_1}}}}} \times 100 = \dfrac{{0.2}}{{3.8}} \times 100 = 0.05\% \dfrac{{[C{H_2} = C = 0]}}{{[C{H_4}] \times 100}}$
So, the maximum percentage yield of the ketene $C{H_2}CO$ obtainable at this temperature is \[5\% \].

Hence, the correct answer is option A.

Note: The alternative method used can be:
Rate of change of Acetic acid
 $ - \dfrac{{d[A]}}{{dt}} = {k_1}[A] + {k_2}[A]$
x = concentration of products formed
So,
 \[x = {[A]^0} - [A]\]
This equation can be used for rate of change of A
 $\dfrac{{dx}}{{dt}} = k1({[A]^0} - x) + k2({[A]^0} - x)$
$\Rightarrow$ $\dfrac{{dx}}{{dt}} = (9k1 + k2)({[A]^0} - X)$
$\Rightarrow$ $(k1 + k2)t = In\left( {\dfrac{{{{[A]}^0}}}{{{{[A]}^0} - x}}} \right)$
$\Rightarrow$ $x = {[A]^0}(1 - {e^{ - (k1 + k2)t}})$
$\Rightarrow$ $x2 = {[A]^0}(1 - {e^{ - (o + k2)t}})$
$\Rightarrow$ $\dfrac{{x2}}{x} = \dfrac{{{{[A]}^0}k2t}}{{{{[A]}^0}(k1 + k2)t}}$
$\Rightarrow$ $\dfrac{{x2}}{x} = \dfrac{{k2}}{{k1 + k2}}$
By solving the equation, we get the maximum percentage yield of ketene.