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QJEEAdv16Chd12f: JEE Advanced 2016 Chemistry - Class 11 JEEAdv16Chd12f Solution: Paragraph Thermal decomposition of gaseous X 2 to gaseous X at…

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Thermal decomposition of gaseous X2 to gaseous X at 298 K takes place according to the following equations:
X2 (g) ⇋\leftrightharpoons 2X (g)
The standard reaction Gibbs energy, ΔrGo\Delta _rG^o , of this reaction is positive. At the start of the reaction, there is one mole of X2 and no X. As the reaction proceeds, the number of moles of X formed is given by β\beta . Thus, βequilibrium\beta _{equilibrium} is the number of moles of X formed at equilibrium. The reaction is carried out at a constant total pressure of 2 bar. Consider the gases to behave ideally. (Given R = 0.083 L bar K-1 mol-1)
Question
The equilibrium constant Kp for this reaction at 298 K, in terms of βequilibrium\beta _{equilibrium} , is

Multiple Choice Options

  • A 8βequilibrium24−βequilibrium2{{8\beta _{equilibrium}^2} \over {4 - {\beta _{equilibrium}^2}}}
  • B 4βequilibrium22−βequilibrium{{4\beta _{equilibrium}^2} \over {2 - {\beta _{equilibrium}}}}
  • C 4βequilibrium24−βequilibrium2{{4\beta _{equilibrium}^2} \over {4 - {\beta _{equilibrium}^2}}}
  • D 8βequilibrium22−βequilibrium{{8\beta _{equilibrium}^2} \over {2 - {\beta _{equilibrium}}}}

Step-by-Step Verified Solution

JEE Advanced 2016 Paper 2 Offline Chemistry - Chemical Equilibrium Question 8 English Explanation

Total number of moles at equilibrium = (1 + α\alpha ) = 1 + (beq / 2)

Therefore,

px=(βeq1+(βeq/2)pT){p_x} = \left( {{{{\beta _{eq}}} \over {1 + ({\beta _{eq}}/2)}}{p_T}} \right)

px2=(1−(βeq/2)1+(βeq/2)pT){p_{{x_2}}} = \left( {{{1 - ({\beta _{eq}}/2)} \over {1 + ({\beta _{eq}}/2)}}{p_T}} \right)

Therefore, the equilibrium constant Kp is given by

Kp=px2px2=(βeq1+(βeq/2)pT)(1−(βeq/2)1+(βeq/2)pT){K_p} = {{p_x^2} \over {{p_{{x_2}}}}} = {{\left( {{{{\beta _{eq}}} \over {1 + ({\beta _{eq}}/2)}}{p_T}} \right)} \over {\left( {{{1 - ({\beta _{eq}}/2)} \over {1 + ({\beta _{eq}}/2)}}{p_T}} \right)}}

=βeq21−(βeq2/4)pT=βeq21−(βeq2/4)×2=8βeq24−βeq2= {{\beta _{eq}^2} \over {1 - (\beta _{eq}^2/4)}}{p_T} = {{\beta _{eq}^2} \over {1 - (\beta _{eq}^2/4)}} \times 2 = {{8\beta _{eq}^2} \over {4 - \beta _{eq}^2}}

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