equilibrium

[匿名]

2008-05-02 2:09 am

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1:

the equation 2IBr(g) = I2(g)+Br2(g)

回答 (1)

Uncle Michael

2008-05-02 4:11 am

✔ 最佳答案

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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