Question

At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO₂ in equilibrium with soild carbon has 90.55% CO by mass
\(C(s) + CO_2(g) ⇋ 2CO(g)\)
Calculate K_c for this reaction at the above temperature.

Answer 1

Given:

Reaction:
C(s) + CO₂(g) ⇋ 2CO(g)

Temperature, T = 1127 K
Total pressure = 1 atm
Composition of gaseous mixture by mass:
CO = 90.55%
CO₂ = 9.45%

---

Step 1: Convert mass percentages into number of moles

Assume total mass of gas mixture = 100 g

Moles of CO =
= 90.55 / 28
= 3.23 mol

Moles of CO₂ =
= 9.45 / 44
= 0.215 mol

Total moles =
= 3.23 + 0.215
= 3.445 mol

---

Step 2: Calculate mole fractions

Mole fraction of CO =
= 3.23 / 3.445
= 0.938

Mole fraction of CO₂ =
= 0.215 / 3.445
= 0.062

---

Step 3: Calculate partial pressures

Total pressure = 1 atm

P_CO = 0.938 atm
P_CO2 = 0.062 atm

---

Step 4: Calculate K_p

K_p = (P_CO)² / P_CO2

K_p = (0.938)² / 0.062

K_p = 0.88 / 0.062

K_p = 14.2

---

Step 5: Calculate K_c

Relation between K_p and K_c:

K_p = K_c(RT)^Δn

For the reaction:
Δn = 2 − 1 = 1

R = 0.0821 L atm mol⁻¹ K⁻¹

RT = 0.0821 × 1127 = 92.5

K_c = K_p / RT

K_c = 14.2 / 92.5

K_c = 0.153

---

Final Answer:

The equilibrium constant at 1127 K is:
K_c = 0.153