NEETChemistryChemical EquilibriumMCQ+4 / −1
The reaction,
2A(g) + B(g) 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
2A(g) + B(g) 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
- A[(0.75)3 (0.25)] [(1.00)2 (1.00)]
- B[(0.75)3 (0.25)] [(0.50)2 (0.75)]
- C[(0.75)3 (0.25)] [(0.50)2 (0.25)]
- D[(0.75)3 (0.25)] [(0.75)2 (0.25)]
View written solutionFree
Correct answer: B
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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