Question

The following equilibrium constants were determined at 1123 K:C(s) + CO2(g) ⇌ 2CO(g) K'P = 1.30 × 1014CO(g) + Cl2(g) ⇌ COCl2(g) K''P = 6.00 × 10-3Calculate the equilibrium constant at 1123 K for the reaction:C(s) + CO2(g) + 2Cl2(g) ⇌ 2COCl2(g)Write the equilibrium constant expression, KP Write the pressures in the following format:(PCO2)

Answer 1

Answer:

4.68x10⁹

Explanation:

Kp is the equilibrium constant based on presure, and depends only on the gas substances. For a generic equation:

aA(g) + bB(g) ⇄ cC(g) + dD(g)

$Kp = \frac{(pC)^c*(pD)^d}{(pA)^a*(pB)^b}$

The reaction given can be summed to form the third one:

C(s) + CO₂(g) ⇄ 2CO (g)  K'p = 1.30x10¹⁴

CO(g) + Cl₂(g) ⇄ COCl₂ (g) K''p = 6.00x10⁻³

We need to multiply the second reaction by 2, so CO will be simplified. If we multiplied a reaction for n, the new Kp will be (Kp)ⁿ, so:

C(s) + CO₂(g) ⇄ 2CO (g)  K'p = 1.30x10¹⁴

2CO(g) + 2Cl₂(g) ⇄ 2COCl₂(g) (K''p)²= (6.00x10⁻³)²

The Kp of the reaction resulted by the sum will be: Kp = K'p*K''p

C(s) + CO₂(g) + 2Cl₂(g) ⇄  2CO(g) + 2COCl₂(g)

Kp = 1.30x10¹⁴ * (6.00x10⁻³)²

Kp = 1.30x10¹⁴*3.60x10⁻⁵

Kp = 4.68x10⁹