Question

The high temperature $\left( \approx 1200K \right)$ decomposition of $C{{H}_{3}}COOH$ which is in the gaseous state occurs as follows, as per simultaneous first order reactions.
\[\begin{align}
& C{{H}_{3}}COOH\xrightarrow{{{K}_{1}}}C{{H}_{4}}+C{{O}_{2}} \\
& C{{H}_{3}}COOH\xrightarrow{{{K}_{2}}}C{{H}_{2}}CO+{{H}_{2}}O \\
\end{align}\]
What will be the % of $C{{H}_{4}}$ by mole in the product mixture if we exclude the $C{{H}_{3}}COOH$?
A. $\dfrac{50{{K}_{1}}}{{{K}_{1}}+{{K}_{2}}}$
B. $\dfrac{100{{K}_{1}}}{{{K}_{1}}+{{K}_{2}}}$
C. $\dfrac{200{{K}_{1}}}{{{K}_{1}}+{{K}_{2}}}$
D. It depends on time

Answer 1

Hint: Think about how the rate constants of the reaction depend on the number of moles of the reactants and the products. Try to relate these and then formulate the answer. Consider the fact that these are first order reactions.

Compete step by step solution:
There is a relation between the number of moles of reactants and the products along with the volume of the whole substance that will undergo the reaction and the rate constant of a reaction. We are going to use this relation to calculate the % of $C{{H}_{4}}$ in the product mixture.
We know that the rate constant of any given reaction is equal to the sum of the number of moles of the product upon the sum of the number of moles of the reactant. This division is further divided by the volume of the reacting substance to get the rate constant. It is expressed as:
\[K=\dfrac{{{n}_{p1}}+{{n}_{p2}}}{{{n}_{r}}\times V}\]
Where, $K$ is the rate constant, ${{n}_{p}}$ are the number of moles of product 1 and product 2, ${{n}_{r}}$ is the number of moles of reactants, and $V$ is the volume of the reactant. We will use this formula to devise the formulae that will be required to calculate ${{K}_{1}}$ and ${{K}_{2}}$. According to this, the formulae are:
\[{{K}_{1}}=\dfrac{{{n}_{C{{H}_{4}}}}+{{n}_{C{{O}_{2}}}}}{{{n}_{C{{H}_{3}}COOH}}\times V}\]
Similarly,
\[{{K}_{2}}=\dfrac{{{n}_{C{{H}_{2}}CO}}+{{n}_{{{H}_{2}}O}}}{{{n}_{C{{H}_{3}}COOH}}\times V}\]
Now, we know that the % of $C{{H}_{4}}$ will be equal to the number of moles of $C{{H}_{4}}$ present divided by the total of the number of moles of all the products. To get an expression similar to this, we will calculate $\dfrac{{{K}_{1}}}{{{K}_{1}}+{{K}_{2}}}$. It will be as follows:
\[\begin{align}
& \dfrac{{{K}_{1}}}{{{K}_{1}}+{{K}_{2}}}=\dfrac{\dfrac{{{n}_{C{{H}_{4}}}}+{{n}_{C{{O}_{2}}}}}{{{n}_{C{{H}_{3}}COOH}}\times V}}{\dfrac{{{n}_{C{{H}_{4}}}}+{{n}_{C{{O}_{2}}}}}{{{n}_{C{{H}_{3}}COOH}}\times V}+\dfrac{{{n}_{C{{H}_{2}}CO}}+{{n}_{{{H}_{2}}O}}}{{{n}_{C{{H}_{3}}COOH}}\times V}} \\
& \dfrac{{{K}_{1}}}{{{K}_{1}}+{{K}_{2}}}=\dfrac{{{n}_{C{{H}_{4}}}}+{{n}_{C{{O}_{2}}}}}{{{n}_{C{{H}_{4}}}}+{{n}_{C{{O}_{2}}}}+{{n}_{C{{H}_{2}}CO}}+{{n}_{{{H}_{2}}O}}} \\
& \dfrac{{{K}_{1}}}{{{K}_{1}}+{{K}_{2}}}=\dfrac{{{n}_{C{{H}_{4}}}}+{{n}_{C{{O}_{2}}}}}{{{n}_{total}}} \\
\end{align}\]
Here, we will assume that ${{n}_{total}}$ is equal to the sum of the number of moles of substance of all the products obtained after dissociation. Now, we know according to the given reaction that the number of moles of carbon dioxide $\left( C{{O}_{2}} \right)$ produced will be equal to the number of moles of methane $\left( C{{H}_{4}} \right)$ produced. So, we can modify the formula to:
\[\begin{align}
& \dfrac{{{K}_{1}}}{{{K}_{1}}+{{K}_{2}}}=\dfrac{2{{n}_{C{{H}_{4}}}}}{{{n}_{total}}} \\
& \dfrac{{{K}_{1}}}{2\left( {{K}_{1}}+{{K}_{2}} \right)}=\dfrac{{{n}_{C{{H}_{4}}}}}{{{n}_{total}}} \\
\end{align}\]
Now, to calculate the percentage, we will multiply both the sides by a factor of 100 and we will get the expression for the value of the % of methane in the product mixture.
\[\begin{align}
& \dfrac{{{n}_{C{{H}_{4}}}}}{{{n}_{total}}}\times 100=\dfrac{{{K}_{1}}}{2\left( {{K}_{1}}+{{K}_{2}} \right)}\times 100 \\
& \dfrac{{{n}_{C{{H}_{4}}}}}{{{n}_{total}}}\times 100=\dfrac{50{{K}_{1}}}{\left( {{K}_{1}}+{{K}_{2}} \right)} \\
\end{align}\]

Hence, the correct answer to this question is option A.

Note: Remember that we do not need to know the volume of the reactant since it is going to get cancelled anyway. We will only require the volume of we need to calculate the actual values of the rate constants and the percentage.