Question

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

Answer 1

Answer:

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$_c$ = $\frac{[NH_3]^2}{[N_2][H_2]^3} = \frac{(0.15)^2}{(0.425)*(0.575)^3}$  = 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.

Answer 2

Answer:

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