Question

At a certain temperature, the equilibrium constant for the following chemical equation is 4.00.

At this temperature, calculate the molarity (M) of NO2(g) that must be added to a starting concentration of 2.73 M SO2(g) (with no products initially present) in order to have formed 1.30 M SO3(g) and 1.30 M NO(g) when equilibrium is reached.

SO2(g)+NO2(g)−⇀↽−SO3(g)+NO(g)

molarity =

Answer 1

SO2 (g) + NO2 (g) <---------------> SO3 (g) + NO (g)

2.73              x                                    0             0 ----------------------> initial

-y                  - y                                 +y           +y --------------------> dissociation

2.73 - y       x - y                                  y           y-----------------------> equilibrium

2.73-1.3      x-1.3                                 1.3          1.3   

1.43           x-1.3                                  1.3          1.3 (afetr below calculation)

at equilibrium SO3 moles given so

y = 1.3

Kc = 4

Kc = [ SO3][NO]/[SO2][NO2]

4= (1.3 )^2 / (1.43) (x - 1.3)

x = 1.6

NO2 moles = x = 1.6

molarity of NO2 = 1.60 M