Question

A high pressure gaseous synthesis of hydrazine (H₄N₂) is governed by the following equilibrium:
`2NH_3 \rightarrow H_4N_2 + H_2`, Keq = 10⁻⁷If a 1 L system is charged with 10 mol of NH₃, how much hydrazine will be present at equilibrium?

Answer 1

The quantity of hydrazine formed is 3.16 mmoles

Step-by-step explanation:

This is a typical equilibrium problem.

Initially we have 10 moles of ammonia, so I don't have any hydrazine or hydrogen. During the reaction, an X amount of moles of ammonia reacts to give half moles of hydrazine and hydrogen. The ratio of reagent to product is 2 to 1.

In equilibrium I will have the 10 moles of ammonia - X and X/2 moles of hydrazine and hydrogen.

2 NH₃ → H₄N₂ + H₂

Initial 10m - -

React x x/2 x/2

Eq 10-x x/2 x/2

Since the volume is 1L the concentration is molar. REMEMBER THAT IN EQUILIBRIUM WE WORK WITH MOLAR CONCENTRATIONS

The expression of Kc is [H₄N₂]. [H2] / [NH3]²

1x10⁻⁷ = [x/2] . [x/2] / (10-x)²

1x10⁻⁷ = 1/4x² / 100-20x+x²

1x10⁻⁷ (100-20x+x²) = 1/4x²

1/4x² - 1x10⁻⁵ + 2x10⁻⁶x = 0 This is a quadratic equation

1/4 = a ; 2x10⁻⁶ = b ; -1x10⁻⁵ = c

(-b +- √(b²-4ac))/2a

x1 = 6.32x10⁻³

x2 = -6.32x10⁻³

Concentrations can not be negative, so we take x1

[H₄N₂] = x/2

[H₄N₂] = 6.32x10⁻³/2 = 3.16x10⁻³ moles

1mmol = 1000 moles

3.16x10⁻³ moles . 1000 = 3.16 mmoles