Given the reaction...#SO_2(g) + NO_2(g) rightleftharpoons SO_3(g) + NO(g)#...for which #K_"eq"=3.75#, what will be the equilibrium concentrations of products and reactants?

Question

Wyatt Atkins · Accepted Answer

#[SO_3(g)]=0.680+0.216=0.896*mol*L^-1#.

#[NO(g)]=0.680+0.216=0.896*mol*L^-1#.

#[SO_2(g)]=0.680-0.216=0.464*mol*L^-1#.

#[NO_2(g)]=0.680-0.216=0.464*mol*L^-1#.

The reaction is followed by the equilibrium: #SO_2(g) + NO_2(g) rightleftharpoons SO_3(g) + NO(g)# And #K_"eq"=3.75=([SO_3(g)][NO(g)])/([SO_2(g)][NO_2(g)])# And if #x*mol*L^-1# #SO_2# reacts................. #K_"eq"=3.75=((0.680+x)(0.680+x))/((0.680-x)(0.680-x))# #K_"eq"=3.75=(x^2+1.360x+0.462)/(x^2-1.360x+0.462)# Thus, #3.75x^2-5.10x+1.73=x^2+1.360x+0.462# And thus, #2.75x^2-6.46x+1.268=0# This has roots at #x=2.13, or x=0.216# (I used the quadratic equation for this solution!) At equilibrium, it is evident that the smallest value is the sole solution that aligns with the initial conditions. #[SO_3(g)]=0.680+0.216=0.896*mol*L^-1#. #[NO(g)]=0.680+0.216=0.896*mol*L^-1#. #[SO_2(g)]=0.680-0.216=0.464*mol*L^-1#. #[NO_2(g)]=0.680-0.216=0.464*mol*L^-1#. I entered these computed values back into the equilibrium expression as a last check: #(0.896)^2/(0.464)^2~=3.75# as required........ Good question; I am stealing it for my A2 class...........Gee, they will howl; especially as some of them will use #x=2.13#.

Mateo Aguilar · Answer

undefined To determine the equilibrium concentrations of products and reactants, you can use the equilibrium expression and the given value of K_eq.

For the reaction: 
SO2(g) + NO2(g) ⇌ SO3(g) + NO(g)

The equilibrium expression is:
K_eq = [SO3] * [NO] / [SO2] * [NO2]

Given that K_eq = 3.75, you can set up an ICE table (Initial, Change, Equilibrium) and use it to solve for the equilibrium concentrations:

Initial concentrations:
[SO2] = initial concentration
[NO2] = initial concentration
[SO3] = 0 (since it starts as a product)
[NO] = 0 (since it starts as a product)

Change:
Let x be the change in concentration.
[SO2] decreases by x
[NO2] decreases by x
[SO3] increases by x
[NO] increases by x

Equilibrium concentrations:
[SO2]_eq = initial concentration - x
[NO2]_eq = initial concentration - x
[SO3]_eq = x
[NO]_eq = x

Now, plug these equilibrium concentrations into the equilibrium expression and solve for x:

3.75 = (x * x) / ([initial concentration - x] * [initial concentration - x])

Solve this quadratic equation for x, then use the equilibrium concentrations to find the numerical values.