Question

A mixture of ` 1*57 " mol of " N_(2) , 1*92 " mol of " H_(2) and 8*13 " mol of " NH_(3) " is introduced into a 20 L reaction vessel at 500 K. At this temperature , the equilibrium constant ," K_(c)` for the reaction ,
`N_(2) (g) + 3 H_(2) (g) hArr 2 NH_(3) " is " 1*7 xx 10^(2)`
Is the reaction mixture at equilibrium ? If not, what is the direction of the net reaction ?

Answer 1

The reaction is : ` N_(2) (g) + 3 H_(2) (g) hArr 2 NH_(3) (g) `
` Q_(c) = ([NH_(3)]^(2))/([N_(2) ][H_(2)]^(3)) = ((8*13)/20"mol"L^(-1))^(2)/(((1*57)/2"mol" L^(-1))((1*92 )/20 "mol" L^(-1))^(3))=2* 38 xx 10^(3) `
As ` Q_(c) != K_(c),` the reaction mixture is not in equilibrium .
As ` Q_(c) gtK_(c)` , the et reaction will be in the backward direction .