10) At constant temperature, a mixture containing 0.500 M N2 and 0.800 M H2 in a reaction vessel reaches equilibrium. At equilibrium the concentration of ammonia is 0.150 M. Calculate the equilibrium constant for this reaction. What kind of problem do you think thisis

The answer to the question is;

The equilibrium constant for the reaction is 0.278

Reversibility.

Explanation:

Initial concentration = 0.500 M N₂ and 0.800 M H₂

N₂ (g) + 3·H₂ (g) ⇔ 2·NH₃ (g)

One mole of nitrogen combines with three moles of hydrogen form 2 moles of ammonia

That is 1 mole of ammonia requires 3/2 moles of H₂ and 1/2 moles of N₂

0.150 M of ammonia requires 3/2×0.150 moles of H₂ and 1/2×0.150 moles of N₂

That is 0.150 M of ammonia requires 0.225 moles of H₂ and 0.075 moles of N₂

Therefore at equilibrium we have

Number of moles of Nitrogen = 0.500 M - 0.075 M = 0.425 M

Number of moles of Hydrogen = 0.800 M - 0.225 M = 0.575 M

Number of moles of Ammonia = 0.150 M

K =  = 0.278

The kind of reaction is a reversible one as the equilibrium constant is greater than 0.01 which as general guide, all components in a reaction with an equilibrium constant between the ranges of 0.01 and 100 will be present when equilibrium is reached and the chemical reaction will be reversible.

The equilibrium constant for this reactions is 0.278

Explanation:

Step 1: Data given

Concentration of N2 = 0.500 M

Concentration of H2 = 0.800 M

At equilibrium the concentration of ammonia is 0.150 M

Step 2: The balanced equation

N2 + 3H2 → 2NH3

Step 3: The initial concentrations

[N2] = 0.500 M

[H2] = 0.800 M

[NH3] = 0M

Step 4: The concentration at equilibrium

For 1 mol N2 we need 3 moles H2 to produce 2 moles NH3

[N2] = (0.500 - X)M

[H2] = (0.800 - 3X)M

[NH3] = 2X M = 0.150 M

X = 0.075 M

[N2] = (0.500 - 0.075) = 0.425 M

[H2] = (0.800 - 3X)M = 0.575 M

[NH3] = 2X M = 0.150 M

Step 5: Calculate Kc

Kc = [NH3]² / [N2][H2]³

Kc = [0.150]² / [0.425][0.575]³

Kc = 0.278

The equilibrium constant for this reactions is 0.278

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