Chemistry problem (Mastering Chemistry)I have no idea how to do this one.?

Question

A system at equilibrium contains I2(g) at a pressure of 0.46 atm and I(g) at a pressure of 0.34 atm . The system is then compressed to half its volume. Part A: Find the pressure of I2 when the system returns to equilibrium.Part B: Find the pressure of I when the system returns to equilibrium.Thanks

Johathan Wehner · Accepted Answer

Equilibrium pressure…..It is important to note that this process occurs at a fixed temperature. This will keep the equilibrium constant the same.2I(g)  I2(g) …………….. K = ?0.34 atm…..0.46 atm ………. initial ….. K = 3.980.64 atm ….0.92 atm ………. when volume is cut in half, P’s momentarily jump to these values-2x …………. +x ………………. changes to re-establish equilibriumK = P(I2) / P²(I)K = (0.46 atm) / (0.34 atm)² = 3.98When the volume is cut in half, the pressures will double.Q = (0.92 atm) / (0.68 atm)² = 1.99Since Q is less than K, the system will shift to the right to re-establish equilibriumK = P(I2) / P²(I)3.98 = (0.92+x) / (0.68-2x)²x = .088P(I2) = 1.01 atmP(I) = 0.504 atmCheck. The new partial pressures should give the value of K.K = P(I2) / P²(I)? = 1.01 atm / (0.504 atm)²? = 3.98 …………………………… The new pressures check out.

Johathan Wehner · Answer

Part A:Consider the original equilibrium :_______________ I₂(g) _____ ⇌ _____ 2I(g) ____ KpOriginal eqm: _ 0.46 atm _________ 0.34 atmKp = P(I)² / P(I₂)Kp = 0.34² / 0.46 …… {1}When the system is compressed to half its volume, the total pressure and both the partial pressures of I₂(g) and I(g) are reduced to half and thus the equilibrium is disturbed.P(I₂) = (0.46 atm) × (1/2) = 0.23 atmP(I) = (0.34 atm) × (1/2) = 0.17 atmIn the equation, there are 1 mole of gaseous molecules on the right, but 2 moles on the left.According to Le Chatelier’s principle, when total pressure decreases, the equilibrium position shifts to the right in order to increase the number of gaseous molecules.______________ I₂(g) _____ ⇌ _____ 2I(g) ____ KpInitial: _______0.23 atm ________ 0.17 atmChange: ______ -y atm __________ +2y atmNew eqm: _(0.23 – y) atm ____ (0.17 + 2y) atmWhen the new equilibrium is established:Kp = (0.17 + 2y)² / (0.23 – y) …… {1}As temperature is kept constant, Kp is unchanged, i.e. {2} = {1}(0.17 + 2y)² / (0.23 – y) = 0.34² / 0.46(4y² + 0.68y + 0.0289) / (0.23 – y) = 0.1156 / 0.460.46 × (4y² + 0.68y + 0.0289) = 0.1156 × (0.23 – y)1.84y² + 0.3128y + 0.013294 = 0.026588 – 0.1156y1.84y² + 0.4284y -0.013294 = 0y = [-0.4284 ± √(0.4284 + 4×1..84×0.13294)] / (2×1.84)y = 0.028 or y = -0.26 (rejected)Pressure of I₂(g) at the new equilibrium = (0.23 – 0.028) atm = 0.202 atm ≈ 0.20 atm (to 2 sig. fig.)====Part B:Pressure of I(g) at the new equilibrium = (0.17 + 2×0.028) atm = 0.226 atm ≈ 0.23 atm (to 2 sig. fig.)

Johathan Wehner

Chemistry problem (Mastering Chemistry)I have no idea how to do this one.?

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