Respuesta :
Answer: The rate of disappearance of [tex]N_2[/tex] is 0.172 M/s
Explanation:
Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.
[tex]N_2(g)+3H_2(g)\rightarrow 2NH_3(g)[/tex]
The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.
Rate in terms of disappearance of [tex]N_2[/tex] = [tex]-\frac{1d[N_2]}{dt}[/tex]
Rate in terms of disappearance of [tex]H_2[/tex] = [tex]-\frac{1d[H_2]}{3dt}[/tex]
Rate in terms of appearance of [tex]NH_3[/tex] = [tex]\frac{1d[NH_3]}{2dt}[/tex]
The rate of formation of ammonia is = 0.345 M/s
[tex]\frac{1d[NH_3]}{dt}=0.345 M/s[/tex]
The rate of disappearance of [tex]N_2[/tex] is one by three times the rate of appearance of [tex]NH_3[/tex]
[tex]-\frac{1d[N_2]}{dt}=\frac{d[NH_3]}{2dt}[/tex]
[tex]-\frac{1d[N_2]}{dt}=\frac{0.345M/s}{2}=0.172M/s[/tex]
Thus the rate of disappearance of [tex]N_2[/tex] is 0.172 M/s
If the rate of formation of ammonia is 0.345 M/s, the rate of disappearance of nitrogen is 0.173 M/s.
Let's consider the balanced equation for the formation of ammonia.
N₂(g) + 3 H₂(g) ⇒ 2 NH₃(g)
The molar ratio of N₂ to NH₃ is 1:2. If the rate of formation of ammonia is 0.345 M/s, the rate of disappearance of nitrogen is:
[tex]\frac{0.345molNH_3}{L.s} \times \frac{1molN_2}{2molNH_3} = \frac{0.173molN_2}{L.s} = 0.173 M/s[/tex]
If the rate of formation of ammonia is 0.345 M/s, the rate of disappearance of nitrogen is 0.173 M/s.
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