Question

At equilibrium, the concentrations are [H2] = 5.0 M, [N2] = 10 M, and [NH3] = 3.0 M. What were the concentrations of nitrogen gas and hydrogen gas that were reacted initially?

Answer 1

`[H_2]_0=0.5M`

`[N_2]_0=8.5M`

Step-by-step explanation:

Hello,

In this case, by computing the equilibrium constant considering the law of mass action of the undergoing chemical reaction we obtain:

`3H_2+N_2 \leftrightarrow 2NH_3`

`K_(eq)=((3.0M)^2)/((5.0M)^3(10M)) =7.2x10^(-3)`

Now, the ICE table turns out:

`\ \ \ \ \ \ \ \ \ \ \ \ 3H_2\ \ \ +\ \ \ \ N_2\ \ \ \ \leftrightarrow \ \ \ \ 2NH_3\\I\ \ \ \ \ \ \ \ \ \ \ [H_2]_0\ \ \ \ \ \ \ [N_2]_0\ \ \ \ \ \ \ \ \ \ \ \ \ \ 0\\C\ \ \ \ \ \ \ \ \ \ \ -3x\ \ \ \ \ \ \ \ \ \ x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2x\\E\ \ \ \ \ \ \ \ [H_2]_0-3x\ \ [N_2]_0-x\ \ \ \ \ \ \ \ \ 2x`

Now, the change "
`x`" due to the reaction is computed via the equilibrium concentration of ammonia as shown below:

`2x=3.0M\\x=1.5M`

Therefore, the initial concentrations result:

`[H_2]_0=5.0M-3(1.5M)=0.5M`

`[N_2]_0=10.0M-1.5M=8.5M`

Best regards.