Question

5 moles of $ {\text{S}}{{\text{O}}_{\text{2}}} $ and 5 moles of $ {{\text{O}}_{\text{2}}} $ are allowed to react. At equilibrium, it was found that $ 60\% $ of $ {\text{S}}{{\text{O}}_{\text{2}}} $ is used up. If the pressure of the equilibrium mixture is one atmosphere, the partial pressure of $ {{\text{O}}_{\text{2}}} $ is:
(A) $ 0.52 $ atm
(B) $ 0.21 $ atm
(C) $ 0.41 $ atm
(D) $ 0.82 $ atm

Answer 1

Hint: A chemical equilibrium is said to exist between the reactants and the products of a reaction when the rate of the forward reaction is equal to the equal of the backward reaction. The moles reacting can be calculated from the stoichiometric coefficients of the reaction. This can be used to find the partial pressure of the reactant using the mole fraction.

Complete step by step solution:
A chemical equilibrium that exists between sulphur dioxide and oxygen in a reaction is as follows:
 $ {\text{S}}{{\text{O}}_{\text{2}}} + \dfrac{1}{2}{{\text{O}}_{\text{2}}} \rightleftharpoons {\text{S}}{{\text{O}}_{\text{3}}} $ , as per the question, the number of moles of the reactants in the beginning is,
5 5 0 (Initial), after the reaction starts, $ 60\% $ of $ {\text{S}}{{\text{O}}_{\text{2}}} $ is used up, hence
2 $ 3.5 $ 3 (equilibrium)
 $ 60\% $ of 5 moles = $ 5 - \left( {\dfrac{{60}}{{100}} \times 5} \right) = 2 $ moles o f $ {\text{S}}{{\text{O}}_{\text{2}}} $ .
According to the stoichiometry of the reaction, for 1 mole of sulphur dioxide, $ \dfrac{1}{2} $ mole of oxygen is required. Hence, for the reaction with $ 60\% $ sulphur dioxide, $ 30\% $ oxygen will be required.
Hence, $ 30\% $ of 5 moles = $ 5 - \left( {\dfrac{{30}}{{100}} \times 5} \right) = 3.5 $ moles of oxygen was consumed till the equilibrium was attained.
The amount of sulphur trioxide formed is = $ 0.6 \times 5 $ = 3 moles.
Hence, the total number of moles of the reactants and the products at equilibrium = $ 3.5 + 3 + 2 $ = $ 8.5 $ moles.
So the mole-fraction of oxygen at equilibrium = $ \dfrac{{3.5}}{{8.5}} $
We know that the partial pressure of a gas = mole fraction of that gas multiplied with the total pressure of the gas mixture.
Therefore, the partial pressure of $ {{\text{O}}_{\text{2}}} $ = $ \dfrac{{3.5}}{{8.5}} \times 1 = 0.41 $ atm.
Hence, the correct answer is option C.

Note:
The partial pressure of a gas in a mixture is defined as the pressure exerted by gas on the walls of a container if it alone occupied the entire volume of the container at the same temperature. The partial pressure of a gas is the measure of the thermodynamic activity of the gas molecules.