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標題:

equilibrium

發問:

A 7.24 g of a sample of IBr is placed in a container with a volume of 0.225dm^3 and is then heated to 500K. Some of the IBr decomposes to I2 and Br2. All three substances are in gas phase at 500K. The system is at equilibrium and the measured pressure of Br2(g) in the system is 3.01atm. Calculate the value of the... 顯示更多 A 7.24 g of a sample of IBr is placed in a container with a volume of 0.225dm^3 and is then heated to 500K. Some of the IBr decomposes to I2 and Br2. All three substances are in gas phase at 500K. The system is at equilibrium and the measured pressure of Br2(g) in the system is 3.01atm. Calculate the value of the equilibrium constant Kp for the following equilibrium at 500K (given 1 atm=101kPa, R=8.31JK^-1mol^-1) 更新: the equation 2IBr(g) = I2(g)+Br2(g)

最佳解答:

PV = nRT PV = (m/M)RT P = mRT/(MV) m = 7.24 g, R = 8.31 J mol-1 K-1, T = 500 K Molar mass of IBr, M = 126.9 + 79.9 = 206.8 g mol-1 V = 0.225 dm3 = 0.225/1000 m3 = 0.000225 m3 Initial pressure of IBr, (PIBr)o = (7.24 x 8.31 x 500) / (206.8 x 0.000225) Pa = (7.24 x 8.31 x 500) / (206.8 x 0.000225 x (101 x 103)) atm = 6.401 atm (PI2)o = (PBr2)o = 0 atm 2IBr(g) ≒ I2 + Br2 Partial pressure ratio IBr : I2 : Br2 = 2 : 1 : 1 Partial pressure of Br2 increased = 3.01 atm Hence, partial pressure of I2 increased = 3.01 atm and partial pressure of IBr decreased = 3.01 x (1/2) atm = 1.505 atm At equilibrium : PIBr = 6.401 - 1.505 = 4.896 atm PI2 = PBr2 = 3.01 atm Kp = PI2 x PBr2 / (PIBr)2 Kp = 3.01 x 3.01 / (4.896)2 Kp = 0.378

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