At 1045K the partial pressures of an equilibrium mixture of H2O (g), H2 (g), and O2 (g) are .040, .0045, and .0030 respectively. Calculate the value of the equilibrium constant Kp at 1045K

2H2O (g) <--> 2H2 (g) + O2 (g)

Kp = 26337.45

Explanation:

The equilibrium constant, Kp describes the ratio of product and reactant concentrations at equilibrium in terms of partial pressures.

For a gas-phase reaction,

aA(g) + bB(g) ⇋ cC(g) + dD(g)

the expression for Kp is:

Kp = (Pc^c * Pd^d) / (Pa^a * Pb^b)

In this reaction;

2H2O (g) ⇋ 2H2 (g)  + O2 (g)

Kp becomes;

Kp = (Pc^c *) / (Pa^a * Pb^b)

Kp = 0.040^2 / (0.0045^2) * (0.0030^1)

Kp = 0.040^2  / (6.075e-8)

Kp = 26337.45

the number of valence electrons for each molecule or ion is shown beneath the structure. remember hydrogen will not have more than two electrons. this hydrogen is part of a covalent bond (sharing two electrons). a double bond here would cause hydrogen to share four electrons with phosphorus.

14 days is the half-life of phosphorus-32