Question

A mixture of $S{{O}_{3}}$, $S{{O}_{2}}$ and ${{O}_{2}}$ gases is maintained at equilibrium in 10 litre flask at a temperature at which ${{K}_{c}}$ for the reaction, $2S{{O}_{2}}(g)+{{O}_{2}}(g)2S{{O}_{3}}(g)$ is $100mo{{l}^{-1}}litre$. At equilibrium,
a) If no.of moles of $S{{O}_{3}}$, $S{{O}_{2}}$ in the flask are same, how many moles of ${{O}_{2}}$ are present 0.1
b) If no.of moles of $S{{O}_{3}}$ in the flask are twice the no.of moles of $S{{O}_{2}}$, how many moles of ${{O}_{2}}$ are present 0.4
c) Both a) and b)
d) None

Answer 1

Hint: Since, the question has a reversible reaction, number of moles of oxygen can be found by the equilibrium constant formula given by ${{K}_{c}}=\dfrac{{{[S{{O}_{3}}]}^{2}}}{{{[S{{O}_{2}}]}^{2}}+{{O}_{2}}}$ and then finding the number of moles in given litres of solution.

Complete step by step solution:
We have the chemical equation in terms of a reversible reaction which shows that the rate of formation is equal to rate of decomposition.
We have studied the physical chemistry part as the equilibrium constant is given the ratio of total concentration of products to that of concentration of reactant side.
Therefore, we can write from the given data as,
${{K}_{c}}=\dfrac{{{[S{{O}_{3}}]}^{2}}}{{{[S{{O}_{2}}]}^{2}}+{{O}_{2}}}$
A) From the given data of this option we can say that $[S{{O}_{3}}]=[S{{O}_{2}}]$ .
Let us consider the moles of sulphur dioxide and sulphur trioxide as ‘n’ [since both have equal concentration]
and m moles of ${{O}_{2}}$
Given that ${{K}_{c}}$=100
At equilibrium in 10 litres of solution we can write the concentrations as,
$\begin{align}
& [S{{O}_{3}}]=\dfrac{n}{10} \\
& [S{{O}_{2}}]=\dfrac{n}{10} \\
& [{{O}_{2}}]=\dfrac{m}{10} \\
\end{align}$
Applying equilibrium constant and solving we get,
$100=\dfrac{{{\left[ {n}/{10}\; \right]}^{2}}}{{{\left[ {n}/{10}\; \right]}^{2}}{{\left[ {m}/{10}\; \right]}^{{}}}}$ $\Rightarrow 100=\dfrac{10}{m}$
Therefore, $m=0.1$
Thus there are 0.1 moles of ${{O}_{2}}$ at equilibrium and thus option A) is correct
Similarly,
B) By the data given in next option, we can say that there are m moles of ${{O}_{2}}$ and n moles of $S{{O}_{2}}$ and moles of $S{{O}_{3}}$ is twice of $S{{O}_{2}}$ that is 2n.
Therefore,
$\begin{align}
& [S{{O}_{3}}]=\dfrac{2n}{10} \\
& [S{{O}_{2}}]=\dfrac{n}{10} \\
& [{{O}_{2}}]=\dfrac{m}{10} \\
\end{align}$
Applying ${{K}_{c}}$ formula,
$100=\dfrac{{{\left[ {2n}/{10}\; \right]}^{2}}}{{{\left[ {n}/{10}\; \right]}^{2}}{{\left[ {m}/{10}\; \right]}^{{}}}}$$\Rightarrow 100=\dfrac{40}{m}$
Thus, m=0.4
Therefore, there are 0.4 moles of ${{O}_{2}}$ at equilibrium.
Therefore, option B) is correct.

Thus the correct answer is option C) Both (a) and (b).

Note: The equilibrium constant can also be written in terms of partial pressures for each of the gases because pressure of a gas is directly proportional to the concentration and is denoted by Kp and takes the exact same form as ${{K}_{c}}$. Thus, the question can be slightly twisted on the basis of this fact.