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Question: The degree of dissociation of \( S{O_3} \) is \( \alpha \) at equilibrium pressure \( {P_0} \) . Wha...

The degree of dissociation of SO3S{O_3} is α\alpha at equilibrium pressure P0{P_0} . What is Kp{K_p} for the reaction 2SO3(g)⇄2SO2(g)+O2(g)2S{O_3}\left( g \right) \rightleftarrows 2S{O_2}\left( g \right) + {O_2}\left( g \right) ?
A) P0α32(1−α)3\dfrac{{{P_0}{\alpha ^3}}}{{2{{\left( {1 - \alpha } \right)}^3}}}
B) P0α3(2+α)(1−α)2\dfrac{{{P_0}{\alpha ^3}}}{{\left( {2 + \alpha } \right){{\left( {1 - \alpha } \right)}^2}}}
C) P0α32(1−α)2\dfrac{{{P_0}{\alpha ^3}}}{{2{{\left( {1 - \alpha } \right)}^2}}}
D) None of the above.

Explanation

Solution

Hint : In the given reaction, Sulfur trioxide is dissociating to form Sulfur dioxide and Oxygen. The total pressure of this reaction is given as P0{P_0} so find the partial pressures of each compound. The formula of Kp{K_p} is the ratio of partial pressures of products to the reactants.

Complete Step By Step Answer:
Given to us is a reaction involving the dissociation of Sulfur trioxide to Sulfur dioxide and Oxygen. The equation for this reaction is written as 2SO3(g)⇄2SO2(g)+O2(g)2S{O_3}\left( g \right) \rightleftarrows 2S{O_2}\left( g \right) + {O_2}\left( g \right)
This is an equilibrium reaction. Let the initial partial pressure of Sulfur trioxide be P. We can now write a table showing the partial pressures of products and reactants initially and at equilibrium.

Let us calculate the total pressure at equilibrium from this table.
Ptotal=2P(1−α)+2Pα+Pα{P_{total}} = 2P\left( {1 - \alpha } \right) + 2P\alpha + P\alpha
By solving, we get Ptotal=2P−2Pα+2Pα+Pα=2P+Pα{P_{total}} = 2P - 2P\alpha + 2P\alpha + P\alpha = 2P + P\alpha
It is already given to us that the total pressure at equilibrium is P0{P_0}
By equating these two, we get P0=P(2+α){P_0} = P\left( {2 + \alpha } \right) and we can write this as P=P02+αP = \dfrac{{{P_0}}}{{2 + \alpha }}
Now we write the formula of Kp{K_p} for the given equilibrium reaction as follows:
Kp=[PSO2]2[PO2][PSO3]2{K_p} = \dfrac{{{{\left[ {{P_{S{O_2}}}} \right]}^2}\left[ {{P_{{O_2}}}} \right]}}{{{{\left[ {{P_{S{O_3}}}} \right]}^2}}}
By substituting the values of partial pressures of each component, we get Kp=(2Pα)2(Pα)(2P(1−α))2{K_p} = \dfrac{{{{\left( {2P\alpha } \right)}^2}\left( {P\alpha } \right)}}{{{{\left( {2P\left( {1 - \alpha } \right)} \right)}^2}}}
On solving, we get Kp=4P3α34P2(1−α)2=Pα3(1−α)2{K_p} = \dfrac{{4{P^3}{\alpha ^3}}}{{4{P^2}{{\left( {1 - \alpha } \right)}^2}}} = \dfrac{{P{\alpha ^3}}}{{{{\left( {1 - \alpha } \right)}^2}}}
We have already found a relation between partial pressure P and total pressure P0{P_0} so let us substitute the value of P from that relation.
Now the above equation becomes Kp=P0α3(2+α)(1−α)2{K_p} = \dfrac{{{P_0}{\alpha ^3}}}{{\left( {2 + \alpha } \right){{\left( {1 - \alpha } \right)}^2}}}
Therefore the correct answer is option B.

Note :
It is to be noted that before writing the formula for Kp{K_p} or drawing the table for partial pressures of the compounds, the reaction equation must be balanced. This is because Kp{K_p} is dependent on the exponential values of the partial pressures.