SO2 + NO2 ⇔ SO3 + NO

Question

Kc = 3.00 for the gas phase reaction SO2 + NO2 ⇔ SO3 + NO

If 1.00 mole of SO2 and 1.00 mole of NO2 are put into an evacuated 2.00 L container and allowed to reach equilibrium, set up the RICE table, calculate x and then use this to enter the equilibrium value for the concentration of SO2(g):

Answer 1

Hello Nori

 

First calculate the molarity of SO₂ and NO₂ from their given # of moles and volume.

 

M_SO2 = 1.00 mol / 2.00 L = 0.500 M SO₂

 

M_NO2 = 1.00 mol / 2.00 L = 0.500 M NO₂

 

Next, set up a RICE table.

 

R:   SO₂         +         NO₂         ⇔         SO₃         +         NO

 

I:  0.500 M               0.500 M                  0                        0

 

C:   -x                          -x                      +x                      +x

 

E: (0.500 M - x)     (0.500 M - x)              x                        x

 

 

K_c = ([SO₃] [NO]) / ([SO₂] [NO₂])  =  3.00

 

(x)(x) / (0.500 - x)(0.500 - x)  =  3.00

 

x² / (0.500 - x)²  =  3.00

 

Take the square root of both sides

 

x / (0.500 - x)  =  1.73

 

x = 1.73 (0.500 - x)

 

x = 0.865 - 1.73x

 

2.73x = 0.865

 

x = 0.865 / 2.73 = 0.317

 

 

[SO₂]_eq = 0.500 - x = 0.500 - 0.317 = 0.183 M