Question

The decomposition of the poisonous gas phosgene is represented by the equation.
COCl2(g)⇌CO(g)+Cl2(g)
Values of KP for this reaction are KP=6.7×10−9 at 99.8∘C and KP=4.44×10−2 at 395∘C.

At what temperature is 10 % dissociated when the total gas pressure is maintained at 1.10 atm ?

Answer 1

Answer –

We are given, reaction –

COCl₂(g) <------> CO(g)+Cl₂(g)

Kp = 6.7*10^-9 , T1 = 99.8 +273 = 372.8 K

T2 = 395 +273 = 668 K , Kp = 4.44*10^-2

We need to calculate temperature when 10 % dissociate

Now first we need to calculate the energy of activation

ln k1/k2 = Ea/R * (1/T2 – 1/T1)

ln 6.7*10^-9 /4.44*10^-2 = Ea / 8.314J.mol^-K^-1 * (1/668 – 1/372.8)

-15.70 * 8.314J.mol^-K^-1 = Ea * (-0.00119)

Ea = 130.585 / 0.00119

       = 1.10*10⁵ J/mol

Now 10 % dissociation means

When, 100 % dissociation , then Kp = 6.7*10^-9

So at 10 % = ?

Kp = 6.7*10^-10

So we need to take k2 = 6.7*10^-10

And T2 = ?

So again using the integrate Arrhenius equation –

ln k1/k2 = Ea/R * (1/T2 – 1/T1)

ln 6.7*10^-9 /6.7*10^-10=   1.10*10⁵ J.mol^-1 / 8.314J.mol^-K^-1 * (1/T2 – 1/372.8)

2.30 = 13250.10 * (1/T2 -0.002682)

1/T2 -0.002682 = 2.30 / 13250.10

                           = 0.0001738

1/T2 = 0.0001738 + 0.002686

        = 0.002856

T2 = 350.11 K

T2 = 350.11 -273

      = 77.11^oC