C5H6O3(g) → C2H6(g) + 3 CO(g)?

Question

Consider the decomposition of the compound C5H6O3 as follows below.

C5H6O3(g) → C2H6(g) + 3 CO(g)

When a 5.75-g sample of pure C5H6O3(g) was sealed into an otherwise empty 2.27 L flask and heated to 237°C, the pressure in the flask gradually rose to 1.90 atm and remained at that value. Calculate K for this reaction.

Answer 1

Kp = P(C2H6) P³(CO) / P(C5H6O3).

The molecular weight of C5H6O3 is 114 g/mole, so the initial pressure was P = nRT/V:

Po = [(5.75 g) / (114 g/mol)] (0.082 L*atm/K*mole) (273 + 237 K) / (2.27 L),

Po = 0.930 atm.

In going to equilibrium, P(C2H6) goes from 0 to x, P(CO) goes from 0 to 3x, and C5H6O3 goes from 0.930 to 0.930-x. At equilibrium the total pressure is 1.90 atm:

Pe = P(C2H6) + P(CO) + P(C5H6O3) = 1.90 atm.

Substituting,

1.90 = x + 3x + (0.930 - x)

Solving for x,

x = P(C2H6) = 0.323 atm

3x = P(CO) = 0.970 atm

(0.930 - x) = P(C5H6O3) = 0.607 atm.

Kp is then, from the equilibrium expression above,

Kp = (0.323 atm) (0.970 atm)³ / (0.607 atm) = 0.486 atm³.