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?

Asked by Pragya Singh | 1 year ago |  88

##### 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 Kc 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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Answered by Abhisek | 1 year ago

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