Verified Solution • 2025–26 Session
moderate
QJEEAdv16Chd12f: JEE Advanced 2016 Chemistry - Class 11 JEEAdv16Chd12f Solution: Paragraph Thermal decomposition of gaseous X 2 to gaseous X at…
Paragraph
Thermal decomposition of gaseous X2 to gaseous X at 298 K takes place according to the following equations:
X2 (g) 2X (g)
The standard reaction Gibbs energy, , 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 . Thus, 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 , is
Thermal decomposition of gaseous X2 to gaseous X at 298 K takes place according to the following equations:
X2 (g) 2X (g)
The standard reaction Gibbs energy, , 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 . Thus, 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 , is
Multiple Choice Options
- A
- B
- C
- D
Step-by-Step Verified Solution

Total number of moles at equilibrium = (1 + ) = 1 + (beq / 2)
Therefore,
Therefore, the equilibrium constant Kp is given by