Part A Consider the reaction A 337L reaction vessel at 950 K

Part A: Consider the reaction:

A 3.37-L reaction vessel at 950. K initially contains 0.219 mol of SO₂ and 0.110 mol of O₂. Calculate the total pressure (in atmospheres) in the reaction vessel when equilibrium is reached.

Part b: Consider the reaction:

2 SO2(g) + O2(g) 2 SO3(g)

Kp = 0.355 at 950. K

Solution

part A

             2 SO2(g)     +         O2(g)      <--->       2 SO3(g)

initial 0.219*0.0821*950/3.37   0.11*0.0821*950/3.37         0 atm

         = 5.06 atm                = 2.546 atm

change      - 2x                        -x                     +2x

equili     5.06-2x                    2.546-x                  2x

   Kp = pSO3^2/pSO2^2*pO2

   0.355 = (2x)^2/((5.06-2x)^2(2.546-x))

x = 1.064


at equilibrium,

pSO3 = 2X = 2*1.064 = 2.128 atm

pSO2 = 5.06-2*1.064 = 2.932 atm

pO2 = 2.546-1.064 = 1.482 atm

total pressure = 2.128+2.932+1.482 = 6.542 atm

part b

kp = pB^2/pA

1.0*10^-4 = (1-2x)^2/x

x = 0.496 atm

at equilibrium

partial pressures of A = x = 0.496 atm

B = 1-2*0.496 = 0.008 atm