Respuesta :
Answer :
The rate of the reaction at t=0 is, [tex]9.0\times 10^{-13}M.s^{-1}[/tex]
The overall order of reaction is, second order reaction.
Explanation :
Rate law : It is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.
The general reaction is:
[tex]A+B\rightarrow C+D[/tex]
The general rate law expression for the reaction is:
[tex]\text{Rate}=k[A]^a[B]^b[/tex]
where,
a = order with respect to A
b = order with respect to B
R = rate law
k = rate constant
[tex][A][/tex] and [tex][B][/tex] = concentration of A and B reactant
Now we have to determine the rate law for the given reaction.
The given balanced equations is:
[tex]O_3(g)+NO(g)\rightarrow O_2(g)+NO_2(g)[/tex]
In this reaction, [tex]O_3[/tex] and [tex]NO[/tex] are the reactants.
The rate law expression for the reaction is:
[tex]\text{Rate}=k[O_2][NO][/tex] ..........(1)
Given :
Rate constant = [tex]k=3.0\times 10^{-5}M^{-1}s^{-1}[/tex]
Concentration of [tex]O_3[/tex] = [tex]5.0\times 10^{-4}M[/tex]
Concentration of [tex]NO[/tex] = [tex]6.0\times 10^{-5}M[/tex]
Now put all the given values in equation 1, we get:
[tex]\text{Rate}=(3.0\times 10^{-5}M^{-1}s^{-1})\times (5.0\times 10^{-4}M)\times (6.0\times 10^{-5}M)[/tex]
[tex]\text{Rate}=9.0\times 10^{-13}M.s^{-1}[/tex]
Thus, the rate of the reaction at t=0 is, [tex]9.0\times 10^{-13}M.s^{-1}[/tex]
Now we have to determine the overall order of reaction.
From the given rate law expression we conclude that,
The order of reaction with respect to [tex]O_3[/tex] is, first order reaction.
The order of reaction with respect to [tex]NO[/tex] is, first order reaction.
Thus, overall order of reaction will be:
Overall order of reaction = 1 + 1 = 2
Thus, the overall order of reaction is second order reaction.
