[rcole23]

2) When 1.0 mol of ammonia gas is injected into a 0.50 L flask, the following reaction proceeds to equilibrium.

2NH3(g) <> N2(g) + 3H2(g)

At equilibrium, 0.30 mol of hydrogen gas is present.

a)   Calculate the equilibrium concentrations of N2(g) and NH3(g) (3 marks)

[NH3]initial = 1.0 mol / 0.50 L = 2.0 mol/L

[H2]equilibrium = .30 mol/L

   2NH3     N2                 +    3H2
Initial concentration (mol/L)   2.0   0.00   0.00
Change in concentration (mol/L)   -2x   +x   +3x
Equilibrium concentration (mol/L)   2.0 mol/L – 2x   X   3x

[3H2]equilibrium= .30 mol/L = 3x

X = .30 mol/L / 3

X = 0.1 mol/L

[NH3]equilibrium = 2.0 mol/L – 2(0.1 mol/L)

[NH3]equilibrium = 2.0 mol/L – 0.2 mol/L

[NH3]equilibrium = 1.8 mol/L

   [N2]equilibrium = x = 0.1 mol/L




b) What is the value of Kc? (2 marks)


Kc= products / recants

= [N2] [H2]3 / [NH3]2

= [0.1][.30]3 / [1.8]2

= .0027 / 3.24

= 8.3 X 10-4

[MrTeo]

When 1.0 mol of ammonia gas is injected into a 0.50 L flask, the following reaction proceeds to equilibrium.
At equilibrium, 0.30 mol of hydrogen gas is present.

[NH3]initial = 1.0 mol / 0.50 L = 2.0 mol/L

[H2]equilibrium = .30 mol/L

You should work with molarity as you did with ammonia, not with moles.

[rcole23]

When 1.0 mol of ammonia gas is injected into a 0.50 L flask, the following reaction proceeds to equilibrium.
At equilibrium, 0.30 mol of hydrogen gas is present.

[NH3]initial = 1.0 mol / 0.50 L = 2.0 mol/L

[H2]equilibrium = .30 mol/L

You should work with molarity as you did with ammonia, not with moles.

I was not sure if I should have converted H2 equilbrium the same way I did with NH3 because it stated that .30 was the amount at equilibrium, but then I noticed it was still in a gas so how about this?

2) When 1.0 mol of ammonia gas is injected into a 0.50 L flask, the following reaction proceeds to equilibrium.

2NH3(g) <> N2(g) + 3H2(g)

At equilibrium, 0.30 mol of hydrogen gas is present.

a)   Calculate the equilibrium concentrations of N2(g) and NH3(g) (3 marks)

[NH3]initial = 1.0 mol / 0.50 L = 2.0 mol/L

[H2]equilibrium = .30 mol / 0.50 L = 0.6 mol/L

                                   2NH3 <>              N2                 +    3H2
Initial concentration (mol/L)       2.0                 0.00                   0.00
Change in concentration (mol/L)     -2x                 +x                     +3x
Equilibrium concentration (mol/L)   2.0 mol/L – 2x       X                      3x

[3H2]equilibrium= .60 mol/L = 3x

X = .60 mol/L / 3

X = 0.2 mol/L


[NH3]equilibrium = 2.0 mol/L – 2(0.2 mol/L)

[NH3]equilibrium = 2.0 mol/L – 0.4 mol/L

[NH3]equilibrium = 1.6 mol/L

   
[N2]equilibrium = x = 0.2 mol/L




b) What is the value of Kc? (2 marks)


Kc= products / recants

= [N2] [H2]3 / [NH3]2

= [0.2][.60]3 / [1.6]2

= .0432 / 2.56

= 0.016875

[MrTeo]

I was not sure if I should have converted H2 equilbrium the same way I did with NH3 because it stated that .30 was the amount at equilibrium, but then I noticed it was still in a gas so how about this?

I don't really understand what you mean, anyway in a homogeneous equilibrium like this you always have to consider molarities of the compounds; while if the equilibrium is heterogeneous solids can be excluded from Guldberg-Waage's law as the volume occupied is irrelevant if compared to gases. Being more precise you should work with mole fractions as they don't change their value with temperature but when you work with equations like this you can approximate.