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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)

Respuesta :

Answer:

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