If, at a given temperature, the equilibrium constant for the reaction H2(g) Cl2(g) 2HCl(g) is Kp, then the equilibrium constant for the reaction HCl(g) (1/2) H2(g) (1/2)Cl2 (g) can be represented as:

Kp = [(PH2/P°)∧(1/2)eq]*[(PCl2/P°)∧(1/2)eq] / (PHCl(g)/P°)eq

Explanation:

∴ Kp = F(T)

⇒ Kp = (PHCl/P°)eq / [(PCl2/P°)eq]*[(PH2/P°)eq]

∴ P° = 1 bar

⇒ Kp = [(PH2/P°)∧(1/2)eq]*[(PCl2/P°)∧(1/2)eq] / (PHCl(g)/P°)eq

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explanation:

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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.