Verified Solution • 2025–26 Session moderate

QAIPMT-2010-Ch-1943: NEET 2010 Chemistry - Class 11 AIPMT-2010-Ch-1943 Solution: The reaction, \(2A(g)+B(g) 3C(g)+D(g)\) is begun with the…

The reaction, 2A(g)+B(g)⇌3C(g)+D(g)2A(g)+B(g)\rightleftharpoons 3C(g)+D(g) is begun with the concentrations of A and B both at an initial value of 1.00 M. When equilibrium is reached, the concentration of D is measured and found to be 0.25 M. The value for the equilibrium constant for this reaction is given by the expression

Multiple Choice Options

  • A [(0.75)3(0.25)]/[(1.00)2(1.00)][(0.75)^3(0.25)]/[(1.00)^2(1.00)]
  • B [(0.75)3(0.25)]/[(0.50)2(0.75)][(0.75)^3(0.25)]/[(0.50)^2(0.75)]
  • C [(0.75)3(0.25)]/[(0.50)2(0.25)][(0.75)^3(0.25)]/[(0.50)^2(0.25)]
  • D [(0.75)3(0.25)]/[(0.75)2(0.25)][(0.75)^3(0.25)]/[(0.75)^2(0.25)]

Step-by-Step Verified Solution

2A(g) | + | B(g) | ⇌ | 3C(g) | + | D(g) Initial mole | 1 | 1 | 0 | 0 At equilibrium | 1 - (2 _ 0.25) = 0.5 | 1 - 0.25 = 0.75 | 3 _ 0.25 = 0.75 | 0.25 Equilibrium constant, K = [ C ] 3 [ D ] [ A ] 2 [ B ] _ K = [ 0.75 ] 3 [ 0.25 ] [ 0.5 ] 2 [ 0.75 ]

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