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Question

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 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)]÷[(0.50)2(0.75)]
The reaction:
2A(g)+B(g)3C(g)+D(g)
Initially 1 10 0
At Eqm.1 - 2x = 1- 0.51 - x =1 -0.25 3x = 0.75 x = 0.25
K=[C]3[D][A]2[B]

K=(0.75)3(0.25)(0.50)2(0.75)

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