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))+3H_(2(g))hArr2NH_(3(g))" is "1.7xx10^(2)`. What is the direction of the net reaction ?

Explanation

` N_2(g) + 3H_2 (g) hArr 2NH_3 `
` Q_c = ([NH_3(g)]^(2))/([N_2(g) ][H_2(g) ]^(3))= ((8.13//20)^(2))/(((1.57)/(20))xx ((1.92)/(20))^(3))`
` " " = 2.38 xx 10 ^(3) " " ( because V = 20 L)`
Given that ` " " K_c = 1.7 xx 10 ^(2)`
Obvioulsy , ` " "Q_c gt K_c`
Therefore , the reaction is not in equilibrium because ` Q_c ne K_c.` It will proceed in the reerse direction.