Calculate Kp Equilibrium Question?

Question

At 25°C, gaseous SO2Cl2 decomposes to SO2(g) and Cl2(g) to the extent that 12.5% of the original SO2Cl2 (by moles) has decomposed to reach equilibrium. The total pressure (at equilibrium) is 0.900 atm. Calculate the value of Kp for this system. ---I keep gettinng the wrong answer on this one. What are you getting?

I checked and .0125 is not correct. Is there some sort of rounding error?

Answer 1

Let X (atm) be the original (partial or total) pressure of SO2Cl2.
..........SO2Cl2 ⇌ SO2 + Cl2
Initial: X ................ 0 ....... 0
At eq: 7X/8 .......... X/8 .... X/8
Hence the total pressure at equilibrium is (in atm):
0.900 = 7X/8 + X/8 + X/8 = 9X/8
X = 0.900*(8/9) = 0.800

The pressures at equilibrium are:
of SO2Cl2: 7X/8 = 0.700 atm
of SO2: X/8 = 0.100 atm, and
of Cl2: X/8 = 0.100 atm.
Kp = (P of Cl2) * (P of SO2) / (P of SO2Cl2) = 0.100 * 0.100/ 0.700 = 0.0143 (atm)

Probably your mistake, and the mistake in the first answer, is that the equilibrium partial pressure of SO2Cl2of 0.700atm has not been used for Kp.

Answer 2

SO2Cl2 --> SO2 + Cl2
number of mole before reaction: x 0 0
number of mole when reacting: x/8 1/1*(x/8) 1/1*(x/8)
number of mole after reaction: 7x/8 x/8 x/8

total pressure = 0.900 = 7x/8 + x/8 + x/8 = 9x/8
x = 0.900*(8/9) = 0.800

then the pressure (after reaction) of SO2Cl2 = 7x/8 = 0.700, the pressure of SO2 = 0.100, the pressure of Cl2 = 0.100

Kp = (P of Cl2) * (P of SO2) / (P of SO2Cl2) = 0.100 * 0.100/ 0.800 = 0.0125

Answer 3

6.65*10^-3