Question

Sulfuryl chloride,SO2Cl2,a compound with very irritating vapors, is used as a reagent in the synthesis of organic compounds. When heated to a sufficiently high temperature it decomposes. Given: SO2Cl2 (g)  SO2 (g) + Cl2 (g) Kc = 0.0450 A sample of 6.70 g SO2Cl2 was placed in a 1.00 L container and heated to 375 oC a. Calculate the concentrations of all species after equilibrium is reached (Assume, in this case that “x” is neg. for computation purposes). Show with approximation if possible.

Answer 1

Given that;

SO2Cl2(g) <===> SO2(g) + Cl2(g) ---> K = 0.045 at 375C

First calcaulte the molarity of SO2Cl2 as follows:

Number of moles of SO₂Cl₂ = m(SO₂CL₂) / M(SO₂Cl₂)
= 6.70g / 134.965 g/mol
= 4.964×10⁻² moles
molarity = number of moles / volume sin L

[SO₂Cl₂]= 4.964×10⁻² mol / 1.00 L
= 4.964×10⁻²M

Now we wil make ICE table as follow:

SO2Cl2(g) <===> SO2(g) + Cl2(g)

I                       4.964×10⁻²M          0          0

C                      -x                                 +x        +x       

E                      4.964×10⁻²M-x       x          x

Now we write the equilibrium constant for this reaction:

K = [SO₂]∙[Cl₂] / [SO₂Cl₂]

0.045 = x² / (4.964×10⁻²M-x)

To solve the above equation we will get

x² + K∙x - K∙c₀ = 0

x = - (K/2) + √( (K²/4) + K∙c₀ )
x= 2.985×10⁻² M

at equilibrium

[SO₂Cl₂] = 4.964×10⁻²M-x           = 4.964×10⁻²M - 2.985×10⁻²M

= 1.979×10⁻²M

[SO₂] = [Cl₂] =x= 2.985×10⁻²M