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Question

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 measured 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)]
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B
[(0.75)3(0.25)]÷[(0.50)2(0.75)]
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C
[(0.75)3(0.25)]÷[(0.50)2(0.25)]
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D
[(0.75)3(0.25)]÷[(0.75)2(0.25)]
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Solution

The correct option is B $$[(0.75)^{3} (0.25)] \div [(0.50)^{2} (0.75)]$$
The reaction:
                                                   $$2A(g)+B(g)\rightleftharpoons 3C(g)+D(g)$$
 Initially 1 1
At Eqm.1 - 2x = 1- 0.51 - x =1 -0.25 3x = 0.75 x = 0.25 
$$K = \dfrac {[C]^{3}[D]}{[A]^{2}[B]}$$

$$K = \dfrac {(0.75)^{3}(0.25)}{(0.50)^{2}(0.75)}$$

Chemistry

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