Question

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Answer 1

n = 0.3 moles of NO2

n = 0.2 moles of H2

V = 1 L

0.14 moles H2

Kc = concentration in molarity

Kp = partial pressure equilibrium

Kc = [products]/[reactants]

KC = [H2O]⁴[NH3]³/[NO2]²[H2]⁵

Lets calculate the concentrations in equilibrium:

0.14 moles of H2 remain, so 0.06 reacted... lets do stoichiometry

Stoichiometry respect to H2 of H2 = -5/5

Stoichiometry respect to H2 of NO2 = -2/5

Stoichiometry respect to H2 of H2O = +4/5

Stoichiometry respect to H2 of 2NH3 = +2/5

So if 0.06 moles of H2 reacted...:

Moles reacted use stochiometry of H2 of H2 = -5/5*0.06 = -0.06 reacted

Moles reacted use stochiometry of H2 of NO2 = -2/5*0.06 = -0.024 reacted

Moles reacted use stochiometry of H2 of H2O = +4/5*0.06 = +0.048 produced

Moles reacted use stochiometry of H2 of 2NH3 = +2/5*0.06 = +0.024 produced

Lets check the concnetraitons in equilibrium:

[H2] = 0.20-0.06 = 0.14

[NO2] = 0.30 -0.024 = 0.276

[H2O] = 0.048

[NH3] = 0.024

Now substitute in Kc

Kc = [H2O]⁴[NH3]³/[NO2]²[H2]⁵

Kc = [0.048]⁴[0.024]³/[0.276]²[0.14]⁵= 0.179

Kc = 0.179

Now Kp

Use partial pressures:

Kp = PH2O⁴*PNH3²/ PNO2²*PH2⁵

P = nRT/V =n*(0.082 atm