A sample of ammonia gas was allowed to come to equilibrium at 400 K.
2NH₃(g)
N₂(g) + 3H₂(g)
At equilibrium,it was found that the concentration of H₂ was 0.0372 M,the concentration of N₂ was 0.0124 M,and the concentration of NH₃ was 0.175 M.What is K_p for this equilibrium? (R = 0.0821 L ∙ atm/(K ∙ mol) )

Question

A sample of ammonia gas was allowed to come to equilibrium at 400 K.
2NH₃(g)

A sample of ammonia gas was allowed to come to equilibrium at 400 K. 2NH<sub>3</sub>(g) N<sub>2</sub>(g) + 3H<sub>2</sub>(g) At equilibrium,it was found that the concentration of H<sub>2</sub> was 0.0372 M,the concentration of N<sub>2</sub> was 0.0124 M,and the concentration of NH<sub>3</sub> was 0.175 M.What is K<sub>p</sub> for this equilibrium? (R = 0.0821 L ∙ atm/(K ∙ mol) ) A) 4.28 B) 2.85 C) 1.70 × 10<sup>-2</sup> D) 1.95 × 10<sup>-8</sup> E) 2.26 × 10<sup>-2</sup>

N₂(g) + 3H₂(g)
At equilibrium,it was found that the concentration of H₂ was 0.0372 M,the concentration of N₂ was 0.0124 M,and the concentration of NH₃ was 0.175 M.What is K_p for this equilibrium? (R = 0.0821 L ∙ atm/(K ∙ mol) )

Quizplus · Accepted Answer

The balanced equation for the reaction is:

2NH3(g) ⇌ N2(g) + 3H2(g)

The equilibrium constant expression for this reaction is:

Kp = (PN2)(PH2)3/(PNH3)2

where P indicates the partial pressure of each species. However, in this problem we are given concentrations, so we need to use the following equation to relate concentrations to partial pressures:

Pi = niRT/V

where Pi is the partial pressure of species i, ni is the number of moles of species i, R is the gas constant, T is the temperature, and V is the volume.

Using this equation, we can calculate the partial pressures of NH3, N2, and H2 at equilibrium:

PNH3 = (0.175 M)(0.0821 L atm/mol K)(400 K) = 5.76 atm
PN2 = (0.0124 M)(0.0821 L atm/mol K)(400 K) = 0.408 atm
PH2 = (0.0372 M)(0.0821 L atm/mol K)(400 K) = 1.22 atm

Substituting these values into the equilibrium constant expression, we get:

Kp = (0.408 atm)(1.22 atm)3/(5.76 atm)2 = 2.26 × 10-2 atm-2

Converting from atm-2 to bar-2 (since Kp is usually reported in bar-2), we get:

Kp = 2.26 × 10-2 bar-2 = 2.26 × 10-2 × 101.325 Pa/bar × (1 bar/100 kPa)2 = 1.70 × 10-2 kPa-2

Finally, converting from kPa-2 to M-2 (since the concentrations are in M), we get:

KS = Kp(RT)Δn = (1.70 × 10-2 kPa-2)(0.0821 L atm/mol K)(400 K)2 = 1.95 × 10-5 M-2

Rounding to two significant figures, we get:

KS = 2.0 × 10-5 M-2

The closest answer choice is E, 2.26 × 10-2, which is the correct answer.

A Sample of Ammonia Gas Was Allowed to Come to Equilibrium