At 700 K, equilibrium constant for the reaction
H2(g)+I2(g) ⇌ 2HI(g)
is 54.8. If 0.5 molL–1 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?
1 Answer
Solution :-
It is given that equilibrium constant Kc for the reaction
H2(g)+I2(g) ⇌ 2HI(g) is 54.8.
Therefore, at equilibrium, the equilibrium constant K’c for the reaction
2HI(g) ⇌ H2(g)+I2(g)
[HI]=0.5 molL-1 will be \( \dfrac{1}{54.8}\)
Let the concentrations of hydrogen and iodine at equilibrium be x molL–1
[H2]=[I2]=x mol L-1
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