At a certain temperature, equilibrium constant (Kc) is 16 for the reaction, `SO_(2)(g) + NO_(2)(g) rarr SO_(3)(g) + NO(g)`If we take one mole each of all the four gases in a one litre container, what would be the equilibrium concentrations of `NO "and" NO_(2)`?

Answer

[NO]=1.6 mol , `[NO_2]` =0.4 mol

Explanation

Initial concentration of each gas =1 mol
Let number of moles of `NO_2` reacting at equilibrium =x
Then, equilibrium concentrations are :
`{:(,SO_2(g)+, NO_2(g) hArr, SO_3(g) +, NO(g)),("Initial conc.",1,1,1,1),("Equi. Conc." ,(1-x),(1-x),(1+x),(1+x)):}`
`K_c=([SO_3][NO])/([SO_2][NO_2])`
or `((1+x)(1+x))/((1-x)(1-x))`=16
or `((1+x)/(1-x))^2=16` or `(1+x)/(1-x)=4`
or 1+x = 4 -4x
or 5x=3 x=3/5=0.6
`therefore` Concentration at equilibrium
[NO]=1+x=1+0.6 = 1.6 mol
`[NO_2]`=1-x =1 -0.6 = 0.4 mol.