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Question

At 700 K, equilibrium constant for the reaction: H2(g)+I2(g)2HI(g) is 64. If 0.5molL1 of HI(g) is present at equilibrium at 700 K, what are the concentration of H2(g) and I2(g) assuming that we initially started with HI(g) and allowed it to reach equilibrium at 700 K?

A
0.0625 M
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B
0.034 M
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C
0.136 M
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D
0.68 M
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Solution

The correct option is A 0.0625 M
Let molL1 be the equilibrium concentration of hydrogen.
Thus the equilibrium concentrations of hydrogen, iodine, and HI are x,x and 0.5M respectively.
The expression for the equilibrium constant is KC=[HI]2[H2][I2].
Substitute values in the above expression.
64=0.5×0.5x2
x2=0.5×0.564 or x=0.0625M
Thus option A is the right answer.

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