For the reaction 2HI(g) <=> H2(g) + I2(g), express the degree of dissociation (alpha) of HI in terms of the equilibrium constant Kp.

(1 + 2*sqrt(Kp))/2
sqrt((1 + 2Kp)/2)
sqrt(2Kp/(1 + 2Kp))
2*sqrt(Kp)/(1 + 2*sqrt(Kp))

Correct answer: 2*sqrt(Kp)/(1 + 2*sqrt(Kp))

Solution

With total moles constant, Kp = [(alpha/2)(alpha/2)]/(1-alpha)² = alpha²/[4(1-alpha)²]. Taking square root, sqrt(Kp) = alpha/[2(1-alpha)], giving alpha = 2*sqrt(Kp)/(1+2*sqrt(Kp)).

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