At `460^(@)C, K_(C)=81` for the reaction, `SO_(2)(g)+NO_(2)(g)hArrNO(g)+SO_(3)(g)`
A mixture of these gases has the following concentrations of the reactants and products:
`[SO_(2)]=0.04M [NO_(2)]=0.04m`
`[NO]-0.30m [SO]=0.3m`
Is the system at equilibrium? If not, in which direction must the reaction proceed to reach equilibrium. What will be the molar concentrations of the four gases at equilibrium?

Answer

`[SO_(2)]=0.034M;[NO_(2)]=0.034M;[NO]=0.306m;[SO_(3)]=0.306M`

Explanation

`SO_(2)(g)+NO_(2)(g)hArrNO(g)+SO_(3)(g) Q_(c)=((0.3)^(2))/((0.04)^(2))=56.25`
`{:(0.04,0.04,0.3,0.3,),(0.04-x,0.04-x,0.3+x,0.3+x,):}` Here, `O_(C)ltK_(C)` hence reaction will proceed in forward direction to reach at state of equilibrium
`K_(C)=((0.3+x)^(2))/((0.04-x)^(2))=81`
`x=0.006`