Question

At a particular temperature 8.0 mol NO2 gas is placed in a 1.0-L container. Over time the NO2 decomposes to NO and O2: 2NO2(g) 2NO(g) + O2(g) At equilibrium the concentration of NO(g) was found to be 3.8 mol/L. What is the the value of K?

Answer 1

`K_"eq"-=1.25`

Explanation:

We address the equilibrium....

`NO_2(g) rightleftharpoons NO(g) + 1/2O_2(g)`

For which...`K_"eq"=([NO(g)][O_2]^(1/2))/([NO_2(g)])`

Initially....`[NO_2(g)]-=8.0*mol*L^-1`...and we say that `x*mol*L^-1` of the stuff reacts...

And so at equilibrium....

`[NO_2(g)]=8-x`; `[NO(g)]=x`; and `[O_2(g)]=x/2`

But we are given that at equilibrium....`[NO(g)]=3.8*mol*L^-1`...and thus...

`[NO_2(g)]=8-x=8-3.8=4.2*mol*L^-1`...and `[O_2(g)]=1.9*mol*L^-1`...and we can now fill in the blanks...

`K_"eq"=([NO(g)][O_2(g)]^(1/2))/([NO_2(g)])=(3.8xxsqrt(1.9))/(4.2)=1.25`