Question

Given, ${{K}_{C}}$ for $2S{{O}_{2}}+{{O}_{2}}\rightleftharpoons 2S{{O}_{3}}$ in a 10 litre flask at certain T is 100 lit/mol. Now, if the equilibrium pressures of $S{{O}_{2}}$ and $S{{O}_{3}}$ are equal, then the mass of ${{O}_{2}}$ present at equilibrium is:
[A] 6.4 g
[B] 12.8 g
[C] 3.2 g
[D] 16 g

Answer 1

Hint: To solve this, firstly write down the equation for the ${{K}_{C}}$ of the given reaction. You can use the ideal gas equation to convert concentration in terms of pressure. Remember to notice that ${{K}_{C}}$ will eventually be in terms of ${{O}_{2}}$ as the other two have constant pressure at equilibrium.

Complete answer:
Here, the reaction given to us is -
     $2S{{O}_{2}}+{{O}_{2}}\rightleftharpoons 2S{{O}_{3}}$
We know that ${{K}_{C}}$ is the equilibrium constant at constant concentration and we express this in terms of partial pressure. For the above reaction, we can write that-
     ${{K}_{C}}=\dfrac{{{\left[ S{{O}_{3}} \right]}^{2}}}{{{\left[ S{{O}_{2}} \right]}^{2}}\left[ {{O}_{2}} \right]}$
Now, we know that the ideal gas equation is, PV = nRT and we also know that concentration is $\dfrac{n}{V}$.
Therefore, we can write the ideal gas equation as P = CRT.
So, in place of concentration, C , we can write $\dfrac{P}{RT}$
Therefore, the above relation becomes-
${{K}_{C}}=\dfrac{\dfrac{P_{S{{O}_{3}}}^{2}}{RT}}{\dfrac{P_{S{{O}_{2}}}^{2}}{RT}\dfrac{{{P}_{{{O}_{2}}}}}{RT}}$
Also, it is given that at equilibrium, pressures of $S{{O}_{2}}$ and $S{{O}_{3}}$ are equal. So,
     ${{K}_{C}}=\dfrac{1}{\dfrac{{{P}_{{{O}_{2}}}}}{RT}}$
Or, ${{K}_{C}}=\dfrac{1}{\left[ {{O}_{2}} \right]}$
Now, the value of ${{K}_{C}}$ is given to us as 100 lit/mol.
Therefore, putting the value in the above equation, we will get concentration of oxygen gas as-
     $\begin{align}
  & 100=\dfrac{1}{\left[ {{O}_{2}} \right]} \\
 & or,\left[ {{O}_{2}} \right]=\dfrac{1}{100}=0.01\text{ mol/lit} \\
\end{align}$
Now, we have to find out the mass of oxygen gas.
The volume of the flask is 10 lit. And, concentration is $\dfrac{n}{V}$.
Therefore, number of moles, n = $0.01\text{ }mol/lit\times 10lit$ = 0.1 mol
We know that the molar mass of a substance is its mass divided by the number of moles.
We know molar mass of ${{O}_{2}}$ is 32 g/mol
Therefore, mass of ${{O}_{2}}$ = $32\text{ g/mol }\times 0.1\text{ mol}$ = 3.2 g

Therefore, the correct answer is option [C] 3.2 g.

Note:
For gases we use equilibrium constant at constant pressure or concentration using the ideal gas equation as ${{K}_{p}}\text{ }and\text{ }{{K}_{c}}$ .
It is important to remember that the value of ${{K}_{p}}\text{ }and\text{ }{{K}_{c}}$ is different for different reactions. The general relation between ${{K}_{p}}\text{ }and\text{ }{{K}_{c}}$ can be written as \[{{K}_{p}}={{K}_{c}}{{\left( RT \right)}^{\Delta n}}\] where the terms have their usual meanings and$\Delta n$is the difference between the total number of moles of reactants and total number of moles in the product.