Answer :
Answer: It would decrease by a factor of 8
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]Rate=k[NO]^2[O_2]^1[/tex] (1)
When the concentration of both reactants were halved:
[tex]Rate'=k{\frac{[NO]}{2}^2}{\frac{[O_2]}{2}^1}[/tex] (2)
Dividing (2) by (1)
[tex]\frac{Rate'}{Rate}=\frac{k\frac{[NO]}{2}^2\times \frac{([O_2]}{2})^1}{k[NO]^2[O_2]^1}[/tex]
[tex]\frac{Rate'}{Rate}=\frac{1}{8}[/tex]
Thus it would decrease by a factor of 8
The reaction rate will:
a. It would decrease by a factor of 8
Rate law:
It describe the relationship between the rate of a chemical reaction and the concentration of its reactants. It states 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.
Rate = [tex]k[NO_2]^2[O_2]^1[/tex]............(1)
When the concentration of both reactants were halved. Then rate will be:
Rate' = [tex]k=[NO_2]^{(1/2)}[O_2]^{(1/2)}[/tex]............(2)
Dividing equation 2 by 1:
[tex]\frac{\text{Rate'}}{\text{Rate}}=\frac{k=[NO_2]^{(1/2)}[O_2]^{(1/2)}}{k[NO_2]^2[O_2]^1} \\\\\frac{\text{Rate'}}{\text{Rate}}=\frac{1}{8}[/tex].
Thus, the rate would decrease by a factor of 8.
Hence, option a is correct.
Find more information about Rate law here:
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