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

? you expect us to answer this? ? this is your schoolwork, i don't think we can with this, i think it would considering being

the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

negative enthalpy of formation indicates that the heat is released during the reaction process. the heat is released from reaction to the surroundings the enthalpy is expressed with a negative sign with the energy given out in the reaction process.

the formation of water takes place by giving out heat which is -285.8 kj/mol.