Answer:
[tex]\frac{1}{-1} \frac{d[A]}{dt} =\frac{1}{-3} \frac{d[B]}{dt}=\frac{1}{1} \frac{d[C]}{dt}=\frac{1}{2} \frac{d[D]}{dt}[/tex]
[tex]\frac{d[A]}{dt} =-0.9\frac{mol}{dm^3*s} \\\\ \frac{d[C]}{dt} =0.9\frac{mol}{dm^3*s}\\\\ \frac{d[D]}{dt} =1.8\frac{mol}{dm^3*s}[/tex]
Explanation:
Hello!
In this case, since the reaction rate for all the species are written by considering how they change and the coefficients they have in the reaction, for A, B, C and D we can write:
[tex]\frac{1}{-1} \frac{d[A]}{dt} =\frac{1}{-3} \frac{d[B]}{dt}=\frac{1}{1} \frac{d[C]}{dt}=\frac{1}{2} \frac{d[D]}{dt}[/tex]
Whereas the numeric denominators stand for the coefficients balancing the reaction. In such a way, by using the rate of consumption of B, we can compute the rate of consumption of A and the rates of formation of both C and D as follows:
[tex]\frac{d[A]}{dt} =\frac{-1}{-3} \frac{d[B]}{dt}=\frac{1}{3}*-2.7\frac{mol}{dm^3*s} \\\\ \frac{d[A]}{dt} =-0.9\frac{mol}{dm^3*s} \\\\\\ \frac{d[C]}{dt} =\frac{1}{-3} \frac{d[B]}{dt}=\frac{1}{-3}*-2.7\frac{mol}{dm^3*s} \\\\ \frac{d[C]}{dt} =0.9\frac{mol}{dm^3*s}\\\\\\ \frac{d[D]}{dt} =\frac{2}{-3} \frac{d[B]}{dt}=\frac{2}{-3}*-2.7\frac{mol}{dm^3*s} \\\\ \frac{d[D]}{dt} =1.8\frac{mol}{dm^3*s}[/tex]
Whereas the rates of consumption are negative and the rates of formation positive.
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