Answer:
0.229 M/s is molecular nitrogen reacting
Explanation:
[tex]N_2(g)+3H_2(g)\rightarrow NH_3(g)[/tex]
Rate of the reaction :
[tex]R=-\frac{1}{1}\frac{d[N_2]}{dt}=-\frac{1}{3}\frac{d[H_2]}{dt}=\frac{1}{2}\frac{d[NH_3]}{dt}[/tex]
Rate at which hydrogen gas is reaction :
[tex]\frac{d[H_2]}{dt}=0.0687 M/s[/tex]
Rate at which hydrogen gas is reaction :
[tex]\frac{d[N_2]}{dt}=?[/tex]
[tex]-\frac{1}{1}\frac{d[N_2]}{dt}=-\frac{1}{3}\frac{d[H_2]}{dt}[/tex]
[tex]\frac{d[N_2]}{dt}=\frac{1}{3}\times 0.0687 M/s[/tex]
[tex]=0.229 M/s[/tex]
0.229 M/s is molecular nitrogen reacting
