Question

5 moles of $S{{O}_{2}}$​ and 5 moles of ${{O}_{2}}$ are allowed to react to form $S{{O}_{3}}$ in a closed vessel. At the equilibrium, 60% of $S{{O}_{2}}$​ is used up. The total number of moles in the vessel at equilibrium is:
(A) 10.0
(B) 8.5
(C) 10.5
(D) 3.9

Answer 1

Hint: When the rate of the forward reaction equals the rate of the backward reaction, a chemical equilibrium between the reactants and products of a reaction is said to exist. The reaction's stoichiometric coefficients can be used to determine the moles that are reacting and that are formed.

Complete Step by Step Solution:
As per the given question, $S{{O}_{2}}$ reacts with ${{O}_{2}}$ to give $S{{O}_{3}}$. The equilibrium reaction for the same can be written as follows:
$S{{O}_{2}}+\dfrac{1}{2}{{O}_{2}}S{{O}_{3}}$

According to the given question 60% of $S{{O}_{2}}$is used up. Initially 5 moles of $S{{O}_{2}}$and 5 moles of ${{O}_{2}}$ are present and reacting with each other.
So, we can say that initially there are 5 moles of$S{{O}_{2}}$, 5 moles of ${{O}_{2}}$ and 0 moles of$S{{O}_{3}}$.

We can see that 1 mole of$S{{O}_{2}}$is reacting with half mole of ${{O}_{2}}$ to give 1 mole of $S{{O}_{3}}$. So, we can conclude that when 60% of $S{{O}_{2}}$is used, 30% of ${{O}_{2}}$would be used.
So at equilibrium:
Moles of$S{{O}_{2}}$ present$=5-5\times 60%$
$=5-5\times 0.6=5-3=2$moles
Moles of${{O}_{2}}$ present$=5-5\times 30%$
$=5-5\times 0.3=5-1.5=3.5$moles

We can see from the reaction that 1 mole of$S{{O}_{2}}$ gives 1 mole of$S{{O}_{3}}$ . So when 60% of $S{{O}_{2}}$used then 60% of$S{{O}_{3}}$ will be formed.
Moles of$S{{O}_{3}}$present$=5\times 60%$
$=5\times 0.6=3$moles
Therefore, total moles present at equilibrium will be equal to the sum total of all the moles of different molecules present at equilibrium.
Then, total moles= $2+3.5+3=8.5$moles
Hence, the correct option is B. 8.5

Note: The most important thing to remember is that we calculate the remaining number of moles of reactant at equilibrium and not the used number of moles of reactants. The moles product formed will depend on the number of moles of reactant used and their stoichiometric coefficients.