Question

Nitrogen reacts with hydrogen to produce ammonia. N2(g) + 3H2(g) 2NH3(g) A mixture of 1.00mol of nitrogen, 3.00mol of hydrogen and 1.98mol of ammonia is allowed to reach equilibrium in a sealed vessel under certain conditions. It was found that 1.64mol of nitrogen were present in the equilibrium mixture. What is the value of Kc under these conditions?

Answer 1

You'll need to make a RICE table..

\(R...~N_2(g)~~~~~+~~~~~3H_2(g)~~~~~⇌~~~~2NH_3(g)\)
\(I...~1.00~mol~~~~~~~~3.00~mol~~~~~~~~~~~~~1.98~mol\)
\(C...~-x~~~~~~~~~~~~~~~-3x~~~~~~~~~~~~~~~~~~~+2x\)
\(E...~1.00-x~~~~~~~~3-3x~~~~~~~~~~~~~~~~~1.98+2x\)


Now, you are given that at equilibrium, there was 1.64 mols of Nitrogen gas. You can use this to find the value of x.
\(1.00-x=1.64\)
\(x=-0.64\)
Now, you can use these values of x to find the mols of Hydrogen and Ammonia at equilibrium.

\(3-3(-0.64)=4.92\)
>>4.92 mols of \(H_2(g)\) at equilibrium
\(1.98+2(-0.64)=0.7\)
>>0.7 mols of \(NH_3(g)\) at equilibrium.

Now that you have found the mols of each gas at equilibrium, you can use this to find the value of \(K_c\).
\(K_c=\frac{[NH_3]^2}{[N_2][H_2]^3}\)

Now the only issue is that you're not given concentration, and you're not given any other additional piece of information like volume to find the concentration. Ideally, you'd need to use concentration and not just mols. But I'll just assume that they are all in a volume of 1 liter so that the mols is the same.

\(K_c=\frac{(0.7)^2}{(1.64)(4.92)^3}\)
Find that and round to nearest sig fig (Ihere it should be 3 sig figs) \(\ddot\smile\)