For the nitrogen fixation reaction 3H2 (g) + N2(g) ↔2NH3 (g), Kc= 6.0 x 10-2 at 500°C. If 0.250 M H2 and 0.050 M NH3 are present at equilibrium, what is the equilibrium concentration of N2?

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llltkl | College Teacher | (Level 3) Valedictorian

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The expression for equilibrium constant, `K_c` for the given reaction is:

`K_c=[NH_3]^2/([H_2]^3[N_2])`

Putting in the equilibrium concentration values,

`6.0*10^(-2)=(0.05)^2/((0.25)^3*[N_2])`

`rArr [N_2]=(0.05)^2/((0.25)^3*6.0*10^(-2))`

`=2.67 M`

Therefore, the equilibrium concentration of `N_2` is 2.67 M.

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