Ahigh pressure gaseous synthesis of hydrazine (h₄n₂) is governed by the following equilibrium: , keq = 10⁻⁷if a 1 l system is charged with 10 mol of nh₃, how much hydrazine will be present at equilibrium?

The quantity of hydrazine formed is 3.16 mmoles

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

a c d and f

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