Respuesta :
Answer: The rate law expression is [tex]\text{Rate}=k[NO]^2[O_2]^1[/tex] and value of 'k' is [tex]1.727\times 10^3M^{-2}s^{-1}[/tex]
Explanation:
Rate law 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.
For the given chemical equation:
[tex]2NO(g)+O_2(g)\rightarrow 2NO_2(g)[/tex]
Rate law expression for the reaction:
[tex]\text{Rate}=k[NO]^a[O_2]^b[/tex]
where,
a = order with respect to nitrogen monoxide
b = order with respect to oxygen
- Expression for rate law for first observation:
[tex]8.55\times 10^{-3}=k(0.030)^a(0.0055)^b[/tex] ....(1)
- Expression for rate law for second observation:
[tex]1.71\times 10^{-2}=k(0.030)^a(0.0110)^b[/tex] ....(2)
- Expression for rate law for third observation:
[tex]3.42\times 10^{-2}=k(0.060)^a(0.0055)^b[/tex] ....(3)
Dividing 1 from 2, we get:
[tex]\frac{1.71\times 10^{-2}}{8.55\times 10^{-3}}=\frac{(0.030)^a(0.0110)^b}{(0.030)^a(0.0055)^b}\\\\2=2^b\\b=1[/tex]
Dividing 1 from 3, we get:
[tex]\frac{3.42\times 10^{-2}}{8.55\times 10^{-3}}=\frac{(0.060)^a(0.0055)^b}{(0.030)^a(0.0055)^b}\\\\2^2=2^a\\a=2[/tex]
Thus, the rate law becomes:
[tex]\text{Rate}=k[NO]^2[O_2]^1[/tex]
Now, calculating the value of 'k' by using any expression.
Putting values in equation 1, we get:
[tex]8.55\times 10^{-3}=k[0.030]^2[0.0055]^1\\\\k=1.727\times 10^3M^{-2}s^{-1}[/tex]
Hence, the rate law expression is [tex]\text{Rate}=k[NO]^2[O_2]^1[/tex] and value of 'k' is [tex]1.727\times 10^3M^{-2}s^{-1}[/tex]
The rate law is the relation between the reaction rate and the concentration of the reactants. The rate law of the reaction is [tex]\rm Rate = k[NO]^{2}[O_{2}]^{1}[/tex] and k is [tex]1.727 \times 10^{3}\;\rm M^{-2}\;s^{-1}[/tex].
What are rate law and k?
The relation of the reaction rate with respect to the concentration of the reactants is given by the rate law equation. While k is the equilibrium constant that is the state of no net change in the reactants and the products of the reaction.
The chemical reaction is given as,
[tex]\rm 2\; NO + O_{2} \rightarrow 2\;NO_{2}[/tex]
The rate law equation is given as,
[tex]\rm Rate = k[NO]^{x}[O_{2}]^{y}[/tex]
Here, order for nitrogen monoxide = a and order for oxygen = b
For the first data set, the rate law expression is:
[tex]8.55 \times 10^{-3} = \rm k(0.030)^{x}(0.0055)^{y}[/tex] .......(equation a)
For the second data set, the rate law expression is:
[tex]1.71 \times 10^{-2} =\rm k(0.030)^{x}(0.0110)^{y}[/tex].......(equation b)
For the third data set, the rate law expression is:
[tex]3.42 \times 10^{-2} = \rm k(0.060)^{x}(0.0055)^{y}[/tex] .......(equation c)
Divide and equate a and b:
[tex]\begin{aligned} \dfrac{1.71 \times 10^{-2}}{8.55 \times 10^{-3}} &= \rm \dfrac{(0.030)^{x}(0.0110)^{y}}{(0.030)^{x}(0.0055)^{y}}\\\\2&= \rm 2^{y}\\\\\rm y &= 1 \end{aligned}[/tex]
Divide and equate a and c:
[tex]\begin{aligned} \dfrac{3.42 \times 10^{-2}}{8.55 \times 10^{-3}} &= \rm \dfrac{(0.060)^{x}(0.0055)^{y}}{(0.030)^{x}(0.0055)^{y}}\\\\2^{2} &= \rm 2^{x}\\\\\rm x& = 2\end{aligned}[/tex]
Hence the rate law equation is, [tex]\rm Rate = k[NO]^{2}[O_{2}]^{1}[/tex]
Substituting values to calculate k as:
[tex]\begin{aligned}8.55 \times 10^{-3} &= \rm k[0.030]^{2}[0.0055]^{1}\\\\\rm k &= 1.727 \times 10^{3}\;\rm M^{-2}\;s^{-1} \end{aligned}[/tex]
Therefore, the rate law is [tex]\rm Rate = k[NO]^{2}[O_{2}]^{1}[/tex] and k is [tex]1.727 \times 10^{3}\;\rm M^{-2}\;s^{-1}.[/tex]
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