Question

At 850.0 K, the value of the equilibrium constant K_p for the hydrazine synthesis reaction below is 0.2300.

N_2(g)+H_2(g)---> N₂H_2(g)

If a vessel contains an initial reaction mixture in which [N₂] = 0.02000 M, [H₂] = 0.03000 M, and [N₂H₂] = 1.000×10^-4 M, what will the [N₂H₂] be when equilibrium is reached?

________M

Answer 1

[N₂H₂] at equilibrium = 5.662 x 10^-3M

Explanation

K_p = 0.2300

K_c = K_p / (R * T)ⁿ

where R = gas constant = 0.0821 L-atm/mol-K

T = temperature = 850 K

n = change in the number of gaseous moles

n = stoichiometric gaseous moles of product - stoichiometric gaseous moles of reactant

n = 1 - (1 + 1)

n = -1

K_c = (0.2300) / [(0.0821 L-atm/mol-K) * (850 K)]^-1

K_c = 16.045

ICE table	N2 (g)	H2 (g)		N2H2 (g)
Initial conc.	0.02000 M	0.03000 M		1.000 x 10-4 M
Change	-x	-x		+x
Equilibrium conc.	0.02000 M - x	0.03000 M - x		1.000 x 10-4 M + x

K_c = [N₂H₂]_eq / [N₂]_eq[H₂]_eq

16.045 = (1.000 x 10^-4 M + x) / [(0.02000 M - x) * (0.03000 M - x)]

Solving for x, x = 5.56 x 10^-3 M

equilibrium concentration of N₂H₂ = 1.000 x 10^-4 M + x

equilibrium concentration of N₂H₂ = 1.000 x 10^-4 M + 5.56 x 10^-3 M

equilibrium concentration of N₂H₂ = 5.662 x 10^-3 M