Answer: The rate of the loss of [tex]O_3[/tex] is 0.52M/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.
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.
[tex]2O_3(g)\rightleftharpoons 3O_2(g)[/tex]
Rate of disappearance of [tex]O_3[/tex] =[tex]-\frac{1d[O_3]}{2dt}[/tex]
Rate of formation of [tex]O_2[/tex] =[tex]+\frac{1d[O_2]}{3dt}[/tex]
[tex]-\frac{1d[O_3]}{2dt}=+\frac{1d[O_2]}{3dt}[/tex]
Rate of formation of [tex]O_2[/tex] = [tex]7.78\times 10^{-1}M/s[/tex]
Thus Rate of disappearance of [tex]O_3[/tex] =[tex]\frac{2d[O_2]}{3dt}=\frac{2}{3}\times 7.78\times 10^{-1}M/s=0.52M/s[/tex]
The rate of appearance of oxygen is 0.52 M/s.
We have the equation of the reaction as;
2 O3(g) → 3 O2(g)
Rate of disappearance of O3 = -1/2d[O3]/dt
Rate of appearance of oxygen = 1/3d[O2]/dt
From the equations above, we can notice that;
d[O3]/dt = 2/3d[O2]/dt
We are told in the question that d[O2]/dt = 7.78 x 10-1 M/s
Hence;
d[O3]/dt = 2/3( 7.78 x 10-1 M/s)
d[O3]/dt = 0.52 M/s
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