Question

2 SO3 (g) → 2 SO2 (g) + O2 (g) A sample of 2.0 mole of SO3 (g) is placed into an evacuated 2.0 L container and heated to 700K. The reaction occurs and the concentration of SO3 (g) as a function of time is shown in the graph below. Sorry the graph will not copy but is shows that SO3 at equilibrium is 0.4 mol/L

a) Determine the equilibrium concentrations of

SO2 (g) and O2 (g)

b) Calculate the value of the Kc and Kp equilibriumconstants for the reaction

Answer 1

a)
initial concentration of SO3 = number of moles/ volume
                                                         = 2 mol / 2L
                                                         = 1 mol/L

2 SO3 (g) → 2 SO2 (g) + O2 (g)
1                            0                0    (initial concentration)
1-2x                     2x               x    (at equilibrium)

But given that equilibrium concentration of SO3 = 0.4 mol/L
so,
1-2x = 0.4
x= 0.3 mol/L

equilibrium concentration of SO2 = 2x = 2*0.3 = 0.6 mol/L
equilibrium concentration of O2 = x = 0.3 mol/L

b)
Kc = [SO2]^2 [O2] / [SO3]^2
      = 0.6^2 *0.3/ {0.4^2}
      = 0.675

use:
Kp = Kc* (RT)^ (delta n)
      = 0.675* (0.0821*700)^ (2+1-2)
      = 0.675* (0.0821*700)^1
      = 0.675* (0.0821*700)
       =38.8