Question

consider the reaction: N_2(g)+3h2 <-------> 2NH3(g). Kp for this reactin at 723K is 4.51x10-5. indicate in each reation is at equilibrium. if not, indicate in which direction the reaction must proceed to reach equilibrium.

a. 100 atm NH_3(g), 30 atm N₂(g), 500 atm H_2(g)

_b. 2.0M NH3(g), 3.0M N2(g) , 5.0 H2(g)

must show work to recieve full credit on hw assignment!

Answer 1

a)

Qp = p(NH3)^2 / p(N2)*p(H2)^3

= (100)^2 / (30*500^3)

= 0.00133

Since Qp is greater than Kp,

reaction must proceed to left to achieve equilibrium

b)

T = 723 K

Δ n = number of gaseous molecule in product - number of gaseous molecule in reactant

Δ n = -2

use:

Kp= Kc (RT)^Δn

4.51*10^-5 = Kc *(0.08206*723.0)^(-2)

Kc = 0.1588

Qc = [NH3]^2 /[N2][H2]^3

= (2.0)^2 / (3.0 * 5.0^3)

= 0.0533

Since Qc is less than Kc,

reaction must proceed to right to achieve equilibrium