Question

# 2. Keq = 2.71*10^-35 for the following reaction at 298K: A(s) + 3B(g) ⇄ C(g) a)...

2. Keq = 2.71*10^-35 for the following reaction at 298K: A(s) + 3B(g) ⇄ C(g) a) Find the equilibrium pressure of C(g) in a container at 298K containing excess of A(s) and initial pressure of 1.57*10^4 bar of B. b) What volume of container is expected to contain a single molecule of compound C (1 atm = 1.013 bar)

In terms of molar concentration

Keq = [C]/[B]3

C = n/V = P/RT = (1.57 x 104 /1.013 atm)/0.08206 atm.L/mol.K x 298 K = 633.78 mol/L

A(s) + 3B(g) ⇄ C(g)

excess 633.78 0 initially

-x -3x +x change

excess 633.78 - 3x x equilibrium

Keq = 2.71 x 10-35 given

Keq = [C]/[B]3 solid concentration is taken as unity

x / (633.78 - 3x)3 = 2.71 x 10-35

x should be <<1 as Keq is very less

x / 633.783 = 2.71 x 10-35

x = 6.899 x 10-27 M

n/V =6.899 x 10-27 = P/RT

P = nRT/V =  6.899 x 10-27 mol/L x 0.08206atm.L/mol.K x 298K = 1.687 x 10-25 atm = 1.71 x 10-25 bar

b)

We have molarity = 6.899 x 10-27M

6.899 x 10-27 moles /L

1 molecule = 1/6.023 x 1023 moles

6.899 x 10-27 moles in 1 L

1 mole in 1/6.899 x 10-27 L

1/6.023 x 1023 moles in    1/6.899 x 10-27 L x1/ 6.023 x 1023 = 240.66 L