🤓 Based on our data, we think this question is relevant for Professor May's class at EKU.

Part A

Step 1

**$\mathbf{\left[}{\mathbf{N}}_{\mathbf{2}}\mathbf{\right]}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{1}\mathbf{.}\mathbf{30}\mathbf{}\mathbf{mol}}{\mathbf{1}\mathbf{}\mathbf{L}}$ = 1.3 M**

**$\mathbf{\left[}{\mathbf{H}}_{\mathbf{2}}\mathbf{\right]}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{1}\mathbf{.}\mathbf{65}\mathbf{}\mathbf{mol}}{\mathbf{1}\mathbf{}\mathbf{L}}$ = 1.65 M**

**$\mathbf{\left[}{\mathbf{NH}}_{\mathbf{3}}\mathbf{\right]}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{0}\mathbf{.}\mathbf{100}\mathbf{}\mathbf{mol}}{\mathbf{1}\mathbf{}\mathbf{L}}$ = 0.1 M**

Step 2

Step 3

0 + 2x = 0.100 M

x = 0.100 M/2

x = 0.05 M

[N_{2}] = 1.30 M - x

[N_{2}] = 1.30 M - 0.05 M

[N_{2}] = 1.25 M

[H_{2}] = 1.65 M - 3x

[H_{2}] = 1.65 M - 3(0.05M)

[H_{2}] = 1.65 M - 0.15 M

[H_{2}] = 1.50 M

Consider the formation of ammonia in two experiments.

(a) To a 1.00-L container at 727°C, 1.30 mol of N _{2} and 1.65 mol of H_{2} are added. At equilibrium, 0.100 mol of NH_{3} is present. Calculate the equilibrium concentrations of N _{2} and H_{2}, and find K_{c} for the reaction:

2NH_{3}(g) ⇌ N_{2}(g) + 3H_{2}(g)

(b) In a different 1.00-L container at the same temperature, equilibrium is established with 8.34 x 10^{−2} mol of NH_{3}, 1.50 mol of N_{2}, and 1.25 mol of H_{2} present. Calculate K_{c} for the reaction:

NH_{3}(g) ⇌ 1/2N_{2}(g) + 3/2H_{2}(g)

What is the relationship between the K_{c} values in parts (a) and (b)? Why aren’t these values the same?