Question

Consider the following reaction starting with standard state conditions at 298 K:

N2O4(g)⇌2NO2(g)             ΔG∘rxn=5.4kJ

Determine the partial pressures of the N2O4 and NO2 at the start of the reaction. In other words, what are each of their initial standard state conditions?

The question is asking for only one value. Please show steps on how to solve.

Answer 1

Solution :-

N2O4(g)   ------- > 2NO2(g)

At the standard state condition the pressure of the N2O4 would be 1 atm

Now lets calculate the equilibrium constant from the delta G

Delta G = 5.4 kJ per mol * 1000 J / 1 k J = 5400 J per mol

Delta G = - RTln K

5400 J per mol = - 8.314 Jper mol K * 298 K * ln K

5400 J per mol / - 8.314 Jper mol K * 298 K = ln K

-2.18 = ln K

Antiln -2.18 = K

0.113 = K

Now using the equilibrium constant lets calculate the partial pressure of the each

K= [ NO2]^2/[N2O4]

0.113 = [2x]^2/[1-x]

0.113 * 1-x = 2x^2

Solving for the x we get x= 0.155

So the partial pressure of the each species are as follows

Partial pressure of N2O4 = 1 atm – 0.155 atm = 0.845 atm

Partial pressure of NO2 = 2x = 2*0.155 atm = 0.31 atm