Question

The equilibrium constant \[{K_p}\] for the reaction, \[2S{O_{2(g)}} + {O_{2(g)}} \rightleftharpoons 2S{O_{3(g)}}\] , is 900\[at{m^{ - 1}}\] at 800K. A mixture containing \[S{O_3}\] and \[{O_2}\] having initial partial pressure of 1 and 2 atm respectively is heated at constant volume to equilibrate. Calculate the partial pressure of each gas at 800K at equilibrium.

Answer 1

Hint: The formula for equilibrium constant of this reaction can be written as \[{K_p} = \frac{{{{[{p_{S{O_3}}}]}^2}}}{{{{[{p_{S{O_2}}}{\text{]}}}^2}{\text{[}}{{\text{p}}_{{O_2}}}]}}\]
As there is no sulfur dioxide gas at initial state, the reaction will go backwards to gain equilibrium state.


Formula used: \[{K_p}=\frac{{{{[{p_{S{O_3}}}]}^2}}}{{{{[{p_{S{O_2}}}{\text{]}}}^2}{\text{[}}{{\text{p}}_{{O_2}}}]}}\]


Complete step by step answer:

We are given that initially, the mixture contains \[S{O_3}\] and \[{O_2}\] in the mixture whose partial pressure is 1 and 2 atm respectively. Then, this mixture is heated. Now, we can see in the reaction that the reaction is reversible that means that it can go both ways. We do not have \[S{O_2}\] at the initial state of the reaction, so the reaction will proceed towards formation of \[S{O_2}\] to attain equilibrium. That means reverse reaction will occur. But we need to find the partial pressure of gases at the equilibrium. So, we will have to use the formula of equilibrium and compare the two states i.e. initial state and final state.

Let’s write the reaction and the partial pressure of gases at initial state and final state.
- Suppose that partial pressure of \[S{O_3}\] will decrease by 2x as it is undergoing decomposition, we are taking 2x for ease of calculation as two molecules of \[S{O_3}\] are present as the product of forward reaction. We can write that its initial partial pressure is 1atm and final partial pressure will be (1-2x) atm.
- We can also see that one molecule of \[{O_2}\] gets generated when 2 molecules of \[S{O_3}\] gets decomposed. So, there will be increase in the partial pressure of \[{O_2}\] gas which can be shown by x. So, at initial state, partial pressure of \[{O_2}\] is 2atm, then we can write that at final state, its partial pressure will be (2+x) atm.
- At initial state, there is no \[S{O_2}\] gas, so its partial pressure will be zero but at equilibrium, its partial pressure can be expressed as 2x atm because there are two molecules of \[S{O_2}\] gets formed.

So, \[2S{O_{2(g)}}{\text{ + }}{{\text{O}}_{2(g)}}{\text{ }} \rightleftarrows {\text{ 2S}}{{\text{O}}_{3(g)}}\]
Initial state: 0 atm 2 atm 1 atm
Final state: 2x atm (2+x) atm (1-2x) atm

Now for the given reaction, we can write the equilibrium constant formula as,
\[{K_p} = \frac{{[{p_{product}}]}}{{[{p_{{\text{reactant}}}}{\text{]}}}}\]
\[{K_p} = \frac{{{{[{p_{S{O_3}}}]}^2}}}{{{{[{p_{S{O_2}}}{\text{]}}}^2}{\text{[}}{{\text{p}}_{{O_2}}}]}}\]

Now for final state, we can write this formula by putting available values as,
\[900 = \frac{{{{[1 - 2x]}^2}}}{{{{[2x{\text{]}}}^2}{\text{[2 + x}}]}}\]

We know that an increase in pressure will be very less compared to 2atm, so we can write (2+x) would be relatively the same as 2 atm. So,
\[900 = \frac{{{{[1 - 2x]}^2}}}{{{{[2x{\text{]}}}^2}{\text{[2}}]}}\]
\[1800 = \frac{{{{[1 - 2x]}^2}}}{{{{[2x{\text{]}}}^2}}}\]

So, we can simplify above equation as,
\[1800 = {\left[ {\frac{{1 - 2x}}{{2x}}} \right]^2}\]
\[1800 = {\left[ {\frac{1}{{2x}} - 1} \right]^2}\]
\[42.43 = \frac{1}{{2x}} - 1\]
So, \[42.43 + 1 = \frac{1}{{2x}}\]
\[43.43 = \frac{1}{{2x}}\]
\[x = \frac{1}{{2(43.43)}}\]
\[x = \frac{1}{{86.86}}\]
\[x = 0.0115atm\]

So we can write the partial pressure of each gas as,
\[{p_{S{O_3}}} = (1 - 2x)atm = (1 - 2(0.0115)) = (1 - 0.0230)atm = 0.977atm\]
\[{p_{{O_2}}} = (2 + x)atm = (2 + 0.0115) = 2.0115atm\]
\[{p_{S{O_2}}} = 2x = 2(0.0115) = 0.0230atm\]

Hence, we can conclude that partial pressure of gases \[{p_{S{O_2}}}\], \[{p_{S{O_3}}}\]and \[{p_{{O_2}}}\] will be 0.0230 atm, 0.977 atm and 2.0115 atm respectively.

Note:

Always remember that as two or more molecules of any gas are produced in the reaction, then while calculating the decrease or increase in their partial pressure make sure that you multiply the increase or decrease by the number of molecules present. e.g. We have multiplied the decrease in partial pressure of sulfur trioxide gas here by 2 as its two molecules are there.