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?

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Answer 1 · 2017-01-22T16:26:24

#[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 equilibrium follows the reaction:

#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)#

And so,

#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!)

Clearly, the lower value is the only solution consistent with the starting conditions. And thus at equilibrium:

#[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#.

Just as a check, I reput these calculated values back into the equilibrium expression:

#(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#.