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)

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)

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

jdjxnbfb

explanation:

jjfnc fdnxnxx

2b2t

explanation:

2b2t