Consider the following reaction: A(g) ~2 B(g), ~ means equilibrium Find the equi

Question

Consider the following reaction: A(g) ~2 B(g), ~ means equilibrium Find the equilibrium partial pressures A and B for each of the following different values of Kp. Assume that the initial partial pressure of B in each case is 1.0 atm and that the initial partial pressure of A is 0.0 atm. Make any appropriate simplifying assumptions. Part A Kp = 1.0 Enter your answers numerically separated by a comma. Express your answer using two significant figures. Part B Kp= 1.8

Explanation / Answer

Reaction equation A(g) ? 2B(g) So the equilibrium partial pressures in atm satisfy the relation: Kp = p(B)²/p(A) According to reaction equation one mole A if formed, when 2 moles of B react away. Assuming ideal gas behavior number of moles and partial pressures are proportional: p(X) = n(X)·R·T/V That means when partial pressures of B decreases by x the partial pressure of A increases by x/2. Let x be th pressure change, when equilibrium has established. The the partial pressures in equilibrium are given by: p(B) = p0 - x = 1 - x p(A) = x/2 Kp = (1 - x)² / (x/2) (1/2)·Kp·x = 1 - 2·x + x² x² - 2·(1 + (Kp/4)) + 1 = 0 => x = (1 + (Kp/4)) - v( (1 + (Kp/4))² - 1 ) or x = (1 + (Kp/4)) + v( (1 + (Kp/4))² - 1 ) Second solution is unfeasible because it would yield x>1 thus a negative pressure for B. (i) Kp = 1.4 x = (1 + (1.4/4)) - v( (1 + (1.4/4))² - 1 ) = 0.443 => p(B) = 1 - 0.443 = 0.557 atm p(A) = 0.443/2 = 0.222 atm (ii) Kp = 2.0×10?4 x = (1 + (2.0×10?4/4)) - v( (1 + (2.0×10?4/4))² - 1 ) = 0.990 => p(B) = 1 - 0.990 = 0.010 p(A) = 0.990/2 = 0.495 (iii) Kp = 2.0×105 Here you can make an simplifying approximation. Great value of Kp indicates that equilibrium is far on the right side of the reaction. So only a negligible small amount of B reacts away until equilibrium has established, i.e. x ist small compared to p0=1 and you can approximate: p(B) = 1 - x ˜ 1 => Kp = 1/(x/2) => x = 2/Kp = 2 / 2.0×105 = 1.0×10?5 => p(B) = 1 atm p(A) = 1.0×10?5 atm

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