- #1

- 80

- 2

- Homework Statement
- Consider the decomposition of the compound C5H6O3 as follows:

C5H6O3(g) --> C2H6(g) + 3CO(g)

When a 5.63-g sample of pure C5H6O3(g) was sealed in an otherwise empty 2.50-L flask and heated to 200°C, the pressure in the flask gradually rose to 1.63 atm and remained at that value. Calculate Kc for this reaction.

- Relevant Equations
- equilibrium expression

My solution:

partial pressure of C5H6O3 = mRT/MV = (5.63 g)(0.08206 L*atm/mol*K)(473 K) / (114.098 g/mol)(2.50 L) = 0.766 atm

equilibrium partial pressure of C5H6O3 = 0.766 - x

equilibrium partial pressure of C2H6 = x

equilibrium partial pressure of CO = 3x

total pressure = 0.766 atm - x + x + 3x = 1.63 atm

x = 0.288 atm

Kp = (0.288 atm)(3*0.288 atm)^3 / (0.766-0.288) = 0.389

Kc = Kp / (RT)^(delta n) = 0.389 / [(0.08206 L*atm/mol*K)(473 K)]^1 = 0.01

My answer was wrong. The correct answer is 6.74*10^-6, and was found by doing all the calculations directly in terms of moles and concentration. Could anyone explain where I went wrong? Thanks.

partial pressure of C5H6O3 = mRT/MV = (5.63 g)(0.08206 L*atm/mol*K)(473 K) / (114.098 g/mol)(2.50 L) = 0.766 atm

equilibrium partial pressure of C5H6O3 = 0.766 - x

equilibrium partial pressure of C2H6 = x

equilibrium partial pressure of CO = 3x

total pressure = 0.766 atm - x + x + 3x = 1.63 atm

x = 0.288 atm

Kp = (0.288 atm)(3*0.288 atm)^3 / (0.766-0.288) = 0.389

Kc = Kp / (RT)^(delta n) = 0.389 / [(0.08206 L*atm/mol*K)(473 K)]^1 = 0.01

My answer was wrong. The correct answer is 6.74*10^-6, and was found by doing all the calculations directly in terms of moles and concentration. Could anyone explain where I went wrong? Thanks.