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  • At 700 K, equilibrium constant for the reaction: H_2 (g) + I_2 (g) 2HI (g) is 54.8. If 0.5 mol L^–1 of HI(g) is present at equilibrium at 700 K, what are the concentration of H_2(g) and I_2(g)

7.15     At 700 K, equilibrium constant for the reaction:

                H_{2}_{(g)}+I_{2}_{(g)}\rightleftharpoons 2HI_{(g)}

                is 54.8. If 0.5 mol L-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 700K?

Answers (1)

best_answer

We have,

The equilibrium constant of the reaction = 54.8

moles of HI = 0.5 mol/L

The given reaction is-

H_{2}_{(g)}+I_{2}_{(g)}\rightleftharpoons 2HI_{(g)}

So, the reverse equilibrium constant is K'_c = 1/K_c

Suppose the concentration of hydrogen and iodine at equilibrium be x

[I_2]=[H_2]=x

Therefore,

K'_c = \frac{[H_2][I_2]}{[HI]} = \frac{x^2}{(.5)^2}
 So, the value of x = \sqrt{\frac{0.25}{54.8}}=0.0675(approx)\ mol/ L  

Posted by

manish

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