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Question

The value of Kp for dissociation of


2HI(g)H2(g)+I2(g)

is 1.84×102. If the equilibrium concentration of H2 is 0.4789 mol litre1, calculate the concentration of HI at equilibrium.

A
3.53 mol lit1
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B
0.53 mol lit1
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C
35.3 mol lit1
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D
0.353 mol lit1
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Solution

The correct option is A 3.53 mol lit1
As Δn=0, Kp=Kc=1.84×102.

Also at equilibrium, [H2][I2]=0.4789M.

The expression for the equilibrium constant becomes Kc=[H2][I2][HI]2.

Substitute values in the above expression.

1.84×102=0.4789×0.4789[HI]2.

[HI]=3.53 mol lit1.

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