Question

Find out the value of equilibrium constant for the following reaction at 298 K.

2NH3(g) + CO2(g)  NH2CONH2 (aq) + H2O (l)

Standard Gibbs energy change, ΔrG 0at the given temperature is –13.6 kJ mol–1

(2 marks)

(ii). At 60°C, dinitrogen tetroxide is 50 per cent dissociated. Calculate the standard free

energy change at this temperature and at one atmosphere. (3 marks)​

Answer 1

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Answer 2

Given info : chemical reaction,

2NH3(g) + CO2(g) ⇒NH2CONH2 (aq) + H2O (l)

Standard Gibbs energy change, ΔrG 0at the given temperature is –13.6 kJ mol–1.

To find : find equilibrium constant,

at 60°C , dinitrogen tetroxide is 50 per cent dissociated. Calculate the standard free energy change at this temperature and at one atmosphere.

solution : using formula,

∆G = -2.303RT logK

here, R = 25/3 J/mol/K, T = 298K and ∆G = -13.6 kJ/mol

so, -13.6 × 10³ J/mol = -2.303 × 25/3 J/mol/K × 298K logK

⇒13600 = 2.303 × 25/3 × 298 logK

⇒2.38 = logK

⇒K = antilog(2.38) = 239.88 ≈ 240

Therefore the equilibrium constant would be 240 (approx)

now, 50% of dinitrogen tetraoxide is dissociated.

from chemical reaction,

N2O4 => 2NO2

at eql 1 - a 2a

so, total = 1 - a + 2a = 1 + a

here a = 0.5

so, total moles = 1 + 0.5 = 1.5

now mole fraction of N2O4 x= (1 - a)/(1 + a) = 0.5/1.5 = 0.33

pressure of N2O4 = x × Pi = 0.33 × 1 atm = 0.33 atm

mole fraction of NO2 = 2a/(1 + a) = 1/1.5 = 0.66 mol

pressure of NO2 = 0.66 atm

equilibrium constant, Kp = p[NO2]²/p[N2O4]

= (0.66)²/(0.33)

= 1.32 atm

so now free energy change, ∆G = -2.303RT logKp

= -2.303 × 8.314 × 333K × log(1.32)

= -768.77 J/mol