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For the gas-phase dissociation A(g) <=> B(g) + C(g), the equilibrium…
Question
For the gas-phase dissociation A(g) <=> B(g) + C(g), the equilibrium constant Kp = 3.0 atm. Starting with 1.0 mol of pure A and allowing equilibrium to be reached at a constant total pressure of 2.0 atm, calculate the degree of dissociation (alpha) of A. Give the answer rounded to four decimal places.
✓ Verified answer: 0.7746checked by our engine — not a guess
Step-by-step solution
Starting with 1 mol A, at degree alpha: moles A = 1-alpha, B = alpha, C = alpha, total = 1+alpha.
Mole fractions give partial pressures (P_tot = 2): p_A = (1-alpha)/(1+alpha)*P, p_B = p_C = alpha/(1+alpha)*P.
Kp = p_B*p_C/p_A = [alpha^2/(1-alpha^2)]*P = 3.
So [alpha^2/(1-alpha^2)]*2 = 3, giving alpha^2/(1-alpha^2) = 1.5, alpha^2 = 1.5 - 1.5 alpha^2, 2.5 alpha^2 = 1.5, alpha^2 = 0.6, alpha = 0.7746.
Final answer0.7746
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