Question

(15 points): Phosgene (COC2) is a highly toxic gas that was used as a chemical warfare agent in World War l. Although it is now banned for that purpose by international agreement, phosgene is still produced for industrial uses (where it must be handled very carefully!) It forms at high temperatures from a mixture of carbon monoxide and chlorine gases: 2. CoB)+ Cl2(8)COCI2(8) Kc 4.95 at 600. K If 7.5 atm worth of each reactant are pumped into a 5.00x103 L reactor at 600. K, what will the molar concentrations of all three substances be at equilibrium?

Answer 1

The ideal gas law is written as

P*V = n*R*T

where P is the pressure of a fixed mass of gas which occupies volume V at temperature T. n is the number of moles of the gas corresponding to the mass of the gas taken.

Therefore,

P = n*R*T/V

=====> P = (n/V)*R*T

We know that n/V = [M] where n is the number of moles of the gas and V is the volume of the gas in L; therefore, [M] is the molar concentration of the gas in mol/L or M.

Hence,

P = [M]*R*T

Initially, we have, 7.5 atm of each reactant; therefore,

7.5 atm = [M]*(0.082 L-atm/mol.K)*(600. K)

======> 7.5 atm = [M]*(49.2 L-atm/mol)

======> [M] = (7.5 atm)/(49.2 L-atm/mol)

======> [M] = 0.1524 mol/L ≈ 0.1524 M.

Set up the ICE chart for the reaction.

CO (g) + Cl₂ (g) <=======> COCl₂ (g)

initial                                   0.1524    0.1524                           -

change                                  -x            -x                                +x

equilibrium                     (0.1524 – x)(0.1524 – x)                     x

Set up the expression for K_c.

K_c = [COCl₂]/[CO][Cl₂]

======> 4.95 = (x)/(0.1524 – x)(0.1524 – x)

======> 4.95 = x/(0.1524 – x)²

======> 4.95*(0.1524 – x)² = x

======> 4.95*(0.02322 – 0.3048x + x²) = x

======> 0.114939 – 1.50876x + 4.95x² = x

======> 4.95x² – 1.50876x + 0.114939 – x= 0

======> 4.95x² – 2.50876x + 0.114939 = 0

Solve the quadratic equation for x.

x = [-(-2.50876) √{(-2.50876)² – 4*4.95*0.114939}]/(2*4.95)

= (2.50876 2.00452)/(2*4.95)

The values of x are

x = 0.45589

or x = 0.0509

x cannot be greater than the initial concentration, 0.1524; hence, the only possible value of x is 0.0509.

Therefore, the equilibrium concentrations are

[CO] = [Cl₂] = (0.1524 – 0.0509) M = 0.1015 M ≈ 0.10 M (ans, correct to 2 sig. figs).

[COCl₂] = 0.0509 M ≈ 0.051 M (ans, correct to 2 sig. figs).