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

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