A mixture of hydrogen and nitrogen gases is placed in a 1.0 L reaction vessel and the reaction is allowed to reach equilibrium at 700K. At equilibrium, it is found that the reaction vessel contains 1.6981 moles of H2, 0.5660 moles of N2, and 0.8679 moles of NH3. 3 H2 (g) + N2 (g) 2 NH3 (g) a. What is the value of Kc at this temperature? b. At this temperature, what is the value of Kp for this reaction?

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Chemistry

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6 October, 02:16

A mixture of hydrogen and nitrogen gases is placed in a 1.0 L reaction vessel and the reaction is allowed to reach equilibrium at 700K. At equilibrium, it is found that the reaction vessel contains 1.6981 moles of H2, 0.5660 moles of N2, and 0.8679 moles of NH3. 3 H2 (g) + N2 (g) 2 NH3 (g) a. What is the value of Kc at this temperature? b. At this temperature, what is the value of Kp for this reaction?

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Lindsey Johnson · Answer 1

a. 0.27 = Kc

b. 8.19*10⁻⁵ = Kp

Explanation:

The reaction is this: 3H₂ (g) + N₂ (g) ⇄ 2NH₃ (g)

As we have the moles of each in the equilibrium and the volume is 1L, we assume the concentrations as molarity.

1.6981 mol/L → H₂

0.5660 mol/L → N₂

0.8679 mol/L → NH₃

Let's make the expression for Kc

Kc = [NH₃]² / [N₂]. [H₂]³

Kc = 0.8679² / 0.5660. 1.6981³

Kc = 0.27

Let's calculate Kp, derivated from Kc

Kp = Kc. (RT) ^Δn where:

Δn is the difference between final moles - initial moles. It is governed by stoichiometry. For this case 2 - (1+3) = - 2

Δn it is always for gases

R is the Ideal gases constant

T is Absolute T°

Let's replace data → 0.27. (0.082. 700K) ⁻² = Kp

8.19*10⁻⁵ = Kp