
Assume that the formation of nitrogen dioxide:
2NO(g)+O2(g)→2NO2(g)
is an elementary reaction. (a) Write the rate law for this reaction. (b) A sample of air at a certain temperature is contaminated with 2.0 ppm of NO by volume. Under these conditions, can the rate law be simplified? If so, write the simplified rate law. (c) Under the conditions described in (b), the half-life of the reaction has been estimated to be 6.4 × 103 min. What would the half-life be if the initial concentration of NO were 10 ppm?
(a)

Interpretation:
The rate law for given reaction has to be written.
Concept Introduction:
Rate law:
The rate of a chemical reaction is directly proportional to the initial concentration of reactants. The rate law for a general reaction is given as
A + B→ productRate = k [A][B]k is the rate consant
Explanation of Solution
Given,
2NO(g) + O2(g) → 2NO2(g)
The rate law is written as rate constant multiplied by product of concentration of nitrogen oxide and oxygen with coefficients raised to the power. The rate law is given as
rate = k [NO]2[O2]
(b)

Interpretation:
The simplified rate law for given reaction when air is contaminated with 2.0ppm of nitrogen oxide has to be written.
Concept Introduction:
Rate law:
The rate of a chemical reaction is directly proportional to the initial concentration of reactants. The rate law for a general reaction is given as
A + B→ productRate = k [A][B]k is the rate consant
Explanation of Solution
Given,
2NO(g) + O2(g) → 2NO2(g)
The rate law is written as rate constant multiplied by product of concentration of nitrogen oxide and oxygen with coefficients raised to the power. The rate law is given as
rate = k [NO]2[O2]
This reaction is appeared as a higher order reaction but actually it follows the second order kinetics, this type of reaction is known as pseudo second order reaction. Because the concentration of oxygen is larger than nitrogen oxide the reaction becomes pseudo order reaction. The simplified rate law is given as
rate = k' [NO]2k' =k[O2]
(c)

Interpretation:
The half-life of nitrogen oxide has to be calculated if the initial concentration of NO is 10ppm.
Concept Introduction:
Half-life:
The time required for half of a reactant to be reduced in a reaction is said to be half-life.
- Half-life of a reaction is represented by the symbol as t1/2.
- Half-life is discovered by Ernst Rutherford's in 1907 from the original term half-life period.
- For the second order reaction t1/2=1k[Ao]
Rate law:
The rate of a chemical reaction is directly proportional to the initial concentration of reactants. The rate law for a general reaction is given as
A + B→ productRate = k [A][B]k is the rate consant
Answer to Problem 20.57QP
The half-life of nitrogen oxide from the given initial concentration of nitrogen oxide is 1.3 × 103min
Explanation of Solution
Given,
2NO(g) + O2(g) → 2NO2(g) t1/2 = 6.4 × 103min[NO] = 10ppm
The rate law is written as rate constant multiplied by product of concentration of nitrogen oxide and oxygen with coefficients raised to the power. The rate law is given as
rate = k [NO]2[O2]
The rate law of a pseudo order reaction is given as
rate = k' [NO]2k' =k[O2]
Half-life for a second order reaction is given as,
t1/2 = 1k[A]0[A]0 is initial concentration of the reactant.
The half-life of nitrogen oxide is determined as,
(t1/2)(t1/2)2 = [(A0)2][(A0)1]6.4×103 min(t1/2)2 = 10 ppm2 ppm(t1/22) = 1.3 × 103min
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Chapter 20 Solutions
Chemistry
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