Question

Calculate the value of the equilibrium constant, Kc , for the equilibrium shown below, if 0.124 moles of NO, 0.0240 mole of H2, 0.0380 moles of N2 and 0.0276 moles of H2O vapor were present in a 2.00 L reaction vessel at equilibrium. 2NO(g) + 2H2(g) <—> N2(g) + 2H2O (g)

6.54
0.352
3.27
0.176

Answer 1

the reaction is

2NO(g) + 2H2(g) <—> N2(g) + 2H2O (g)

Kc = [N2] [ H2O]^2 / [NO]^2 [ H2]^2

Given

moles of NO = 0.124 therefore [NO] = moles /volume = 0.124 /2 = 0.062

moles of H2 = 0.0240 , therefore [H2] = moles / volume = 0.0240 / 2 = 0.012

moles of N2 = 0.0380 , therefore [N2] = moles / volume = 0.0380 / 2 = 0.019

moles of H2O = 0.0276 , therefore [H2O] = moles / volume = 0.0276 / 2 = 0.0138

Kc = (0.019) ( 0.0138)^2 / (0.062)^2 ( 0.012)^2 = 6.54