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 $x=2.13$ .

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