Find the grams of the reactants at equilibrium and Find the initial grams of ammonia

Question

Given Nitrogen gas + Hydrogen gas react to make ammonia at EQ. Kc and molarity of AMMONIA are given at equilibrium. You have to use the Kp concert to Kc and the given molarity to find X in a rice box. N2 + 3H2 <> 2NH3 Kp = 4.34 x 10^-3 in 1L at 300C There is 1.05g NH3 in EQ mixture a. Find the grams of the reactants at equilibrium b. Find the initial grams of ammonia

Answer 1

N₂ + 3H₂ <==> 2NH₃ .. balanced equation

Since they ask for initial NH₃ present, we must assume that the reaction starts with NH₃ and it then decomposes to N₂ and H₂. So, we can rewrite the reaction as follows:

2NH₃ <==> N₂ + 3H₂ and then the Kp will be 1 / 4.34x10^-3 = 230.4

Now, we convert Kp to Kc as follows:

Kp = Kc(RT)^∆n where ∆n = change in number of moles of gas = 4 - 2 = 2

230.4 = Kc(0.0821 x 573)²

230.4 = 2213Kc

Kc = 0.104

Set up the RICE table:

2NH₃ <==> N₂ + 3H₂ .................Reaction

i...................0......0.......................Initial

-2x.............+x.....+3x....................Change

i-2x..............x......3x.....................Equilibrium

and i-2x = 1.05 g NH₃ = 1.05 g x 1 mol / 17.03 g = 0.0617 mols

Kc = 0.104 = [N₂][H₂]³ / [NH₃]²

0.104 = (x)(3x)² / (0.0617)²

27x⁴ = 3.96x10^-4

x = 1.10 mols

Equilibrium grams of reactants:

N₂ = x = 1.10 mols x 28 g / mol = 30.8 g N₂

H₂ = 3x = 1.10 mol x 3 = 3.30 mols x 2 g / mol = 6.60 g H₂

Initial grams NH₃ = i

i - 2x = 1.05 g

2x = 2.20 mols

2.20 mols x 17 g / mol = 37.4 g

i - 37.4 g = 1.05 g

i = 38.5 g NH₃ initially present