Answer:
[tex][H_2]_0=0.5M[/tex]
[tex][N_2]_0=8.5M[/tex]
Explanation:
Hello,
In this case, by computing the equilibrium constant considering the law of mass action of the undergoing chemical reaction we obtain:
[tex]3H_2+N_2 \leftrightarrow 2NH_3[/tex]
[tex]K_{eq}=\frac{(3.0M)^2}{(5.0M)^3(10M)} =7.2x10^{-3}[/tex]
Now, the ICE table turns out:
[tex]\ \ \ \ \ \ \ \ \ \ \ \ 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[/tex]
Now, the change "[tex]x[/tex]" due to the reaction is computed via the equilibrium concentration of ammonia as shown below:
[tex]2x=3.0M\\x=1.5M[/tex]
Therefore, the initial concentrations result:
[tex][H_2]_0=5.0M-3(1.5M)=0.5M[/tex]
[tex][N_2]_0=10.0M-1.5M=8.5M[/tex]
Best regards.
