Question

When sulphur is heated at 900 K, ${S_8}$ is converted to ${S_2}$. What will be the equilibrium constant for the reaction if initial pressure of 1 atm falls by 25 % at equilibrium ?
a.) 0.75 $at{m^3}$
b.) 2.55 $at{m^3}$
c.) 25.0 $at{m^3}$
d.) 1.33 $at{m^3}$

Answer 1

Hint: The equilibrium constant is the result of dividing concentration of product by reactant raised to the power of their stoichiometric coefficients. The reaction of conversion of ${S_8}$ into ${S_2}$ is given as -
${S_8} \rightleftharpoons 4{S_2}$
So, ${K_P}$ = $\dfrac{{{{({P_{{S_2}}})}^4}}}{{{P_{{S_8}}}}}$

Complete step by step answer:
Let us first write what is given to us and what we need to find out. This way we can easily solve the question.
So, the things given to us are :
Temperature = 900 K
Initial pressure of ${S_8}$ = 1atm
Things to find out :
Equilibrium constant for the reaction if initial pressure of 1 atm falls by 25 %
So, The reaction is written as -
${S_8} \rightleftharpoons 4{S_2}$
We have the initial pressure of ${S_8}$ = 1 atm
And the initial pressure of ${S_2}$ = 0
At equilibrium,
The pressure of ${S_8}$ falls by 25 %.
So, the pressure of ${S_8}$ = 1 - 0.25 atm
So, the pressure of ${S_8}$ = 0.75 atm
And the pressure of ${S_2}$ = 4 $ \times $0.25
the pressure of ${S_2}$ = 1 atm
We know that the equilibrium constant is the result of dividing concentration of product by reactant raised to the power of their stoichiometric coefficients.
So, ${K_P}$ = $\dfrac{{{{({P_{{S_2}}})}^4}}}{{{P_{{S_8}}}}}$
${K_P}$ = $\dfrac{{{{(1)}^4}}}{{0.75}}$
${K_P}$ = $\dfrac{1}{{0.75}}$
${K_P}$ = 1.33 $at{m^3}$

So, the option d.) is the correct answer.

Note: It must be noted that the equilibrium is a point where the rate of forward reaction is equal to the rate of backward reaction. The amount of reactant that converts into product at equilibrium is equal to the amount of products that converts into reactant back.