Ch4 and h2o were mixed in a 0.64 l reactor at 1800 k. steam reforming took place according to: ch4 (g) + h2o (9) co (g) + 3 h2 (9) the equilibrium constant for this reaction is k+0.28. at equilibrium, the reactor contained 0.36 mol of co, 0.081 mol of ha and 0.051 mol of ch. what is the concentration of h20 at equilibrium?

0.05110 mol/L is the concentration of water at equilibrium.

Explanation:

An equilibrium constant of the given reaction =

Volume of the container = V = 0.64 L

Concentration =

Concentration of CO=

Concentration of

Concentration of

Concentration of

An expression of an equilibrium constant  is given as:

0.05110 mol/L is the concentration of water at equilibrium.

"standard molar enthalpy:

h2o ( liquid water ) : - 285.8 kj/mole

h2o ( water vapor ) : - 241.8 kj / mole.

δhrxn = - 241. 8 - ( - 285.8 )= -241.8 + 285.8 = 44.0

answer: 44.0 kj"

b.final