Question

Ammonia is among the top ten synthesized compounds. Its many uses include the manufacture of ammonium nitrate and other fertilizers. Ammonia decomposes at high temperatures. In an experiment to explore this behavior, 2.00 moles of gaseous NH3 are sealed in a rigid 1-liter vessel. The vessel is heated at 800 K and some of the NH3 decomposes in the following reaction:
2 NH3 (g) <-> N2 (g) + 3 H2 (g)The system eventualky reaches equilibrium and is found to contain 1.740 mole of NH3. What are the values of Kp and Kc for the decomposition reaction at 800k?

Answer 1

Answer:

Kc = 2.55x10⁻³

Kp = 10.97

Explanation:

Based on the reaction:

2 NH₃(g) ⇄ N₂(g) + 3H₂(g)

2 moles of ammonia produce 1 mole of nitrogen and 3 of hydrogen.

Kc is defined as:

Kc = [N₂] [H₂]³ / [NH₃]²

When 2.00 moles of NH₃ are added in a vessel and the reaction occurs, when the reaction is in equilibrium the moles of each specie is:

NH₃: 2.00moles - 2X

N₂: X

H₂: 3X

Where X represents reaction coordinate.

As moles in equilibrium of NH₃ are 1.740moles:

1.740 mol = 2.00 - 2X

0.13moles = X

That means moles of N₂ are 0.13 and H₂ 0.39

Replacing in Kc formula:

Kc = [0.13] [0.39]³ / [1.74]²

Kc = 2.55x10⁻³

To obtain Kp from Kc you need to use the formula:

Kp = Kc (RT)^Δn

Where R is 0.082atmL/molK, T is temperature in Kelvin (800K) and Δn is change in moles of gas, that is moles of products - moles of reactants ( 4 - 2 = 2)

Replacing:

Kp = 2.55x10⁻³(0.082atmL/molKₓ800K)²7

Kp = 10.97

Answer 2

The values of Kp and Kc for the decomposition reaction of ammonia at 800k is 2.5 × 10⁻³ and 10.97 respectively.

What is Kp and Kc?

Kp and Kc both are equilibrium constants with respect to the partial pressure and concentration in molaity respectively.

Given chemical reaction with ICE table will be represented as:

                          2 NH₃(g)  ⇄  N₂(g)  +  3H₂(g)

Initial:                        2                0            0

Change:                  -2x              +x          +3x

Equilibrium:            2-2x             x             3x

According to the question moles of NH₃ at equilibrium is 1.740 mol, so

2 - 2x = 1.740 mol

x = 0.13moles

Kc equation for the given reaction is:

Kc = [N₂].[H₂]³ / [NH₃]²

Kc = (x).(3x)³ / (0.13)²

Kc = 2.5 × 10⁻³

Kp will be calculated as:

Kp =  Kc.(RT)ⁿ, where

R = universal gas constant = 0.082atmL/mol.K

T = temperature = 800K

n = change in moles = 4 -2 = 2

Kp = (2.5 × 10⁻³)[(0.082)(800)]²

Kp = 10.97.

Hence value of Kp & Kc is 2.5 × 10⁻³ and 10.97 respectively.

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