Respuesta :
Answer: A) Only I
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]NO_2(g)+CO(g)\rightarrow NO(g)+CO_2(g)[/tex]
Given:
[tex]Rate=k[NO_2]^2[CO]^1[/tex]
k= rate constant
Order with respect to [tex]NO_2[/tex] =2
Order with respect to [tex]CO[/tex] = 1
I. 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]Rate=-\frac{1d[NO_2]}{2dt}=-\frac{1d[CO]}{dt}[/tex]
[tex]-\frac{1d[NO_2]}{dt}=-2\times \frac{1d[CO]}{dt}[/tex]
Thus the rate of change of [tex]NO_2[/tex] is twice the rate of change of [tex]CO[/tex].
II. Doubling the concentrations of [tex]NO_2[/tex] and [tex]CO[/tex] simultaneously will increase the rate of the reaction by a factor of four
[tex]Rate'=k[2NO_2]^2[2CO]^1[/tex]
[tex]Rate'=k[2]^2[NO_2]^2[2]^1[CO]^1[/tex]
[tex]Rate'=k\times 8[NO_2]^2[2]^1[CO]^1[/tex]
[tex]Rate'= 8\times Rate[/tex]
Thus the rate increases by a factor of 8 and not 4.
Thus the correct statement is only the rate of change of [tex]NO_2[/tex] is twice the rate of change of [tex]CO[/tex].
The fact is that none of the statement above is true or holds reaction of NO2(g) + CO(g) → NO(g) + CO2(g).
Though, when you double the concentration, the rate is said to doubles also. The rate is said to be proportional to the square of the concentration of a specified reactant. And as such, When you double the concentration the rate goes up four times but this is not the case with this equation above.
The rate of reaction is known to be calculated through the use of the formula rate = Δ[C]/Δt. Where Δ[C] is said to be the change in product concentration in course of time Δt.
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