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ChemMedMain 2026 • 5th Apr - Shift 2

Calculate Equilibrium Constant Kp for Gas Dissociation

The reaction A(g)⇌B(g)+C(g)\text{A}(g) \rightleftharpoons \text{B}(g) + \text{C}(g) was initiated with the amount 'aa' of A(g)\text{A}(g). At equilibrium it is found that the amount of A(g)\text{A}(g) remaining is (a−x)(a-x) at a total pressure of pp.

The equilibrium constant KpK_p of the reaction can be calculated from the expression :

Options

A

x2a2+x2×p\frac{x^2}{a^2+x^2} \times p

B

x2a2−x2×p\frac{x^2}{a^2-x^2} \times p

Correct
C

a+x2x2×p\frac{a+x^2}{x^2} \times p

D

a2−x2x2×p\frac{a^2-x^2}{x^2} \times p

Topics & Concepts

EquilibriumChemical Equilibrium

Step-by-Step Solution

To calculate the equilibrium constant KpK_p for the given reaction, we analyze the reaction stoichiometry and mole balance at equilibrium.

The given gas phase dissociation reaction is: A(g)⇌B(g)+C(g)\text{A}(g) \rightleftharpoons \text{B}(g) + \text{C}(g)

Step 1: Determine the moles of each species at equilibrium

  • Initial moles:

    • A(g)=a\text{A}(g) = a
    • B(g)=0\text{B}(g) = 0
    • C(g)=0\text{C}(g) = 0

  • At equilibrium:

    • Moles of A(g)\text{A}(g) remaining = a−xa - x
    • Moles of A(g)\text{A}(g) reacted = xx
    • Moles of B(g)\text{B}(g) formed = xx
    • Moles of C(g)\text{C}(g) formed = xx

Step 2: Calculate total moles at equilibrium (ntotaln_{\text{total}})

ntotal=nA+nB+nC=(a−x)+x+x=a+xn_{\text{total}} = n_{\text{A}} + n_{\text{B}} + n_{\text{C}} = (a - x) + x + x = a + x

Step 3: Calculate the partial pressure of each gas

Using the total pressure pp:

  • pA=(nAntotal)p=(a−xa+x)pp_{\text{A}} = \left( \frac{n_{\text{A}}}{n_{\text{total}}} \right) p = \left( \frac{a - x}{a + x} \right) p
  • pB=(nBntotal)p=(xa+x)pp_{\text{B}} = \left( \frac{n_{\text{B}}}{n_{\text{total}}} \right) p = \left( \frac{x}{a + x} \right) p
  • pC=(nCntotal)p=(xa+x)pp_{\text{C}} = \left( \frac{n_{\text{C}}}{n_{\text{total}}} \right) p = \left( \frac{x}{a + x} \right) p

Step 4: Calculate the equilibrium constant KpK_p

The equilibrium constant expression is: Kp=pB⋅pCpAK_p = \frac{p_{\text{B}} \cdot p_{\text{C}}}{p_{\text{A}}}

Substituting the partial pressure values: Kp=(xa+x⋅p)(xa+x⋅p)(a−xa+x⋅p)K_p = \frac{\left( \frac{x}{a + x} \cdot p \right) \left( \frac{x}{a + x} \cdot p \right)}{\left( \frac{a - x}{a + x} \cdot p \right)}

Kp=x2(a+x)2⋅p2a−xa+x⋅pK_p = \frac{\frac{x^2}{(a + x)^2} \cdot p^2}{\frac{a - x}{a + x} \cdot p}

Kp=x2(a+x)(a−x)×pK_p = \frac{x^2}{(a + x)(a - x)} \times p

Kp=x2a2−x2×pK_p = \frac{x^2}{a^2 - x^2} \times p

Therefore, the correct option is B.

Calculate Equilibrium Constant Kp for Gas Dissociation | Chemistry PYQ Solution - JEE Challenger