Answer:
[tex]r_{N_2}=-0.1725M/s[/tex]
Explanation:
Hello,
In this case, by means of the law of mass action, we firstly write the described chemical reaction:
[tex]N_2(g)+3H_2(g)\rightarrow 2NH_3(g)[/tex]
Thus, as ammonia is being formed at 0.345 M/s, nitrogen will be disappearing at (consider law of mass action):
[tex]r_{NH_3}=0.345M/s\\\\\frac{1}{-1} r_{N_2}=\frac{1}{-3}r_{H_2}=\frac{1}{2} r_{NH_3}\\\\r_{N_2}=-\frac{1}{2} r_{NH_3}=-\frac{1}{2} *0.345M/s\\\\r_{N_2}=-0.1725M/s[/tex]
Best regards.
