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Chemical Equilibrium question

2010 · Q65
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Chemical Equilibrium question

2010 · Q65

NEETChemistryChemical EquilibriumMCQ+4 / −1
The reaction,
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 measuread and found to be 0.25 M. The value for the equilibrium constant for this reaction is given by the expression
  1. A
    [(0.75)3 (0.25)] ÷\div÷ [(1.00)2 (1.00)]
  2. B
    [(0.75)3 (0.25)] ÷\div÷ [(0.50)2 (0.75)]
  3. C
    [(0.75)3 (0.25)] ÷\div÷ [(0.50)2 (0.25)]
  4. D
    [(0.75)3 (0.25)] ÷\div÷ [(0.75)2 (0.25)]
View written solutionFree

Correct answer: B

2A(g)+B(g)⇌3C(g)+D(g)
Initial mole1100
At equilibrium1 - (2 $ \times $ 0.25)
= 0.5
1 - 0.25
= 0.75
3 $ \times $ 0.25
= 0.75
0.25


Equilibrium constant, K = $${{{{\left[ C \right]}^3}\left[ D \right]} \over {{{\left[ A \right]}^2}\left[ B \right]}}$$

$ \therefore $ K = $${{{{\left[ {0.75} \right]}^3}\left[ {0.25} \right]} \over {{{\left[ {0.5} \right]}^2}\left[ {0.75} \right]}}$$
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