Question

When sulphur (in the form of $ {S_8} $ ) is heated to temperature T, at equilibrium, the pressure of S8 falls by 30% from 1.0 atm, because $ {S_8}(g) $ is partially converted into $ {S_2}(g) $ .Find the value of $ {K_p} $ for this reaction.
(A) 2.96
(B) 6.14
(C) 204.8
(D)None of these

Answer 1

Hint :Chemical equilibrium is a state where the rate of forward reaction becomes equal to the rate of backward reaction in case of reversible reactions. It is of dynamic nature. Concentration of the reactants and products becomes constant at Equilibrium. Equilibrium constant is the value of reaction quotient found as the ratio of Concentration of products and that of reactants. In a gaseous reaction where we have partial pressure then it is calculated as ratio of Concentration in terms of partial pressure of the products and the reactants $ ({K_p}) $ .

Complete Step By Step Answer:
Consider a reaction, $ aA(g) + bB(g) \rightleftharpoons cC(g) + dD(g) $
The expression for $ {K_p} = {\dfrac{{{{\left[ {{P_C}} \right]}^c}\left[ {{P_D}} \right]}}{{{{\left[ {{P_A}} \right]}^a}{{\left[ {{P_B}} \right]}^b}}}^d} - - - - (1) $
 $ {K_p} $ is related to Equilibrium constant in terms of concentration as-
 $ {K_p} = {K_c}{\left( {RT} \right)^{\Delta n}} - - - - (2) $
Where $ \Delta n $ is the difference in the number of moles of products and reactants.
In the question we are given a reaction as follows,
 $ {S_8}(g) \rightleftharpoons 4{S_2}(g) $

At time t=0	1	0
Partial pressure at Equilibrium	1-x	4x

{Since at time t=0, we have initial partial pressure of 1 atm and product partial pressure will be 0,now after some time reactant starts changing into product. The partial pressure of reactant decreases by an amount of x so the partial pressure of product will increase by 4x since 4 is the stoichiometric coefficient}
Given in the question, partial pressure falls by 30% from 1 atm means x=30/100=0.30
Now calculating $ {K_p} = \dfrac{{{{\left[ {4x} \right]}^4}}}{{1 - x}} = \dfrac{{{{\left[ {4 \times 0.30} \right]}^4}}}{{1 - 0.30}} = \dfrac{{{{\left[ {1.2} \right]}^4}at{m^4}}}{{0.7atm}} = 2.96at{m^3} $
Hence option (A) is correct.

Note :
 $ {K_p} $ has the unit of pressure (i.e. atm or bar). $ {K_p} $ can be calculated in Homogeneous as well as Heterogeneous Equilibria. For Example calculating $ {K_p} $ in Haber’s process. It is important to note that Equilibrium constant depends on the mode of representation of reaction, temperature, change in pressure or the change in concentration of the reactants and product.