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Question

At 700 K, the equilibrium constant for the reaction;
H2(g)+I2(g)2HI(g) is 54.8.

If 0.5 mol litre1 of HI(g) is present at equilibrium at 700 K, what are the concentrations of H2(g) and I2(g) assuming that we initially started with HI(g) and allowed it to reach equilibrium at 700 K.

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Solution

H2(g)+I2(g)2HI(g)
Kc=[HI]2[H2][I2]=54.8

[H2][I2]=[HI2]Kc

[HI]=0.5 mol litre1;
[H2][I2]=[0.5]254.8=4.56×103
At equilibrium, equal concentration of H2 and I2 will exist if dissociation of HI is carried out. Thus,

[H2]=[I2]

[H2]2=4.56×103
[H2]=0.068 mol litre1

[I2]=0.068 mol litre1

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