Explanation:
As the given reaction is [tex]3A + B + 2C \rightarrow D + E[/tex]
Also, rate = k[tex][A]^{2}[C]^{2}[/tex]
So, rate is zero order with respect to B and it is second order with respect to both A and C.
It is given that [A] = [tex]2.4 \times 10^{-6} M[/tex] and [tex][A]_{0} = [tex]2.20 \times 10^{-5} M[/tex]
For second order rate law, equation is as follows.
[tex]\frac{1}{[A]} = \frac{1}{[A]_{0}} + kt[/tex]
[tex]\frac{1}{2.4 \times 10^{-6}} = \frac{1}{2.20 \times 10^{-5}} + k \times 7 min[/tex]
k = [tex]0.412 \times 10^{5} M^{-1} sec^{-1}[/tex]
As, the initial rate = k [tex][A]^{2}_{o} [C]^{2}_{o}[/tex]
= 0.412 \times 10^{5} M^{-1} sec^{-1} \times (2.20 \times 10^{-5} M)^{2} \times (0.6 M)^{2}[/tex]
= [tex]0.717 \times 10^{-5} M^{3} sec^{-1}[/tex]
Thus, we can conclude that the initial rate of the reaction described above is [tex]0.717 \times 10^{-5} M^{3} sec^{-1}[/tex].
