Nitrogen (II) oxide, \({\rm{NO}}\), reacts with hydrogen, \({{\rm{H}}_{\rm{2}}}\), according to the following equation: What would the rate law be if the mechanism for this reaction were:

\({\rm{2NO + 2}}{{\rm{H}}_{\rm{2}}}{\rm{ }} \to {\rm{ }}{{\rm{N}}_{\rm{2}}}{\rm{ + 2}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\)

What would the rate law be if the mechanism for this reaction were: \(\begin{aligned}{l}{\rm{2NO + }}{{\rm{H}}_{\rm{2}}}{\rm{ }} \to {\rm{ }}{{\rm{N}}_{\rm{2}}}{\rm{ + }}{{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{2}}}{\rm{ (slow)}}\\{{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{2}}}{\rm{ + }}{{\rm{H}}_{\rm{2}}}{\rm{ }} \to {\rm{ 2}}{{\rm{H}}_{\rm{2}}}{\rm{O (fast)}}\end{aligned}\)

Short Answer

Expert verified

The rate law for the first reaction mechanism would be,

\({\bf{rate = k(NO}}{{\bf{)}}^{\bf{2}}}{{\bf{(}}{{\bf{H}}_{\bf{2}}}{\bf{)}}^{\bf{2}}}\)

and

The rate law for the second reaction mechanism would be,

\({\rm{rate = }}k{{\rm{(NO)}}^2}{\rm{(}}{{\rm{H}}_{\rm{2}}}{\rm{)}}\)

Step by step solution

01

Definition of Rate Equation (Rate Law)

The rate law (or rate equation) is the mathematical expression which explains thethe relationship between the rate of a chemical reaction and the concentration of its reactants.

\({\rm{rate = }}k{{\rm{(A)}}^x}{{\rm{(B)}}^y}{{\rm{(C)}}^z}.....\)

Where,

(A), (B), and (C) denotes the molar concentrations of reactants.

kis the rate constant.

Exponents m, n, and pare generally positive integers.

02

Rate Law for the First Reaction 

\({\rm{2NO + 2}}{{\rm{H}}_{\rm{2}}}{\rm{ }} \to {\rm{ }}{{\rm{N}}_{\rm{2}}}{\rm{ + 2}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\)

For the first reaction with mechanism,

\({\rm{2NO + 2}}{{\rm{H}}_{\rm{2}}}{\rm{ }} \to {\rm{ }}{{\rm{N}}_{\rm{2}}}{\rm{ + 2}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\)

The rate law will be,

\({\bf{rate = k(NO}}{{\bf{)}}^{\bf{2}}}{{\bf{(}}{{\bf{H}}_{\bf{2}}}{\bf{)}}^{\bf{2}}}\)

03

Rate Law for the Second Reactions

\(\begin{aligned}{\rm{2NO + }}{{\rm{H}}_{\rm{2}}}{\rm{ }} \to {\rm{ }}{{\rm{N}}_{\rm{2}}}{\rm{ + }}{{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{2}}}{\rm{ (slow)}}\\{{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{2}}}{\rm{ + }}{{\rm{H}}_{\rm{2}}}{\rm{ }} \to {\rm{ 2}}{{\rm{H}}_{\rm{2}}}{\rm{O (fast)}}\end{aligned}\)

For the second reaction with mechanism,

\(\begin{aligned}{\rm{2NO + }}{{\rm{H}}_{\rm{2}}}{\rm{ }} \to {\rm{ }}{{\rm{N}}_{\rm{2}}}{\rm{ + }}{{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{2}}}{\rm{ (slow)}}\\{{\rm{H}}_{\rm{2}}}{{\rm{O}}_{\rm{2}}}{\rm{ + }}{{\rm{H}}_{\rm{2}}}{\rm{ }} \to {\rm{ 2}}{{\rm{H}}_{\rm{2}}}{\rm{O (fast)}}\end{aligned}\)

The rate law is determined by depending on its slow reaction. The slow step in the reaction mechanism is considered as rate determining step. Thus, the rate law will be

\({\bf{rate = k(NO}}{{\bf{)}}^{\bf{2}}}{\bf{(}}{{\bf{H}}_{\bf{2}}}{\bf{)}}\)

Therefore, the rate law will be\({\bf{rate = k(NO}}{{\bf{)}}^{\bf{2}}}{{\bf{(}}{{\bf{H}}_{\bf{2}}}{\bf{)}}^{\bf{2}}}\)and\({\bf{rate = k(NO}}{{\bf{)}}^{\bf{2}}}{\bf{(}}{{\bf{H}}_{\bf{2}}}{\bf{)}}\)for both the reactions with different mechanisms.

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Most popular questions from this chapter

Both technetium-99 and thallium-201 are used to image heart muscle in patients with suspected heart problems. The half-lives are 6 h and 73 h, respectively. What percent of the radioactivity would remain for each of the isotopes after 2 days (48 h)?

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Time (s)

9.0

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(P) (M)

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Some bacteria are resistant to the antibiotic penicillin because they produce penicillinase, an enzyme with a molecular weight of \({\bf{3 \times 1}}{{\bf{0}}^{\bf{4}}}\)g/mole that converts penicillin into inactive molecules. Although the kinetics of enzyme-catalysed reactions can be complex, at low concentrations this reaction can be described by a rate equation that is first order in the catalyst (penicillinase) and that also involves the concentration of penicillin. From the following data: 1.0 L of a solution containing 0.15 µg (\({\bf{0}}{\bf{.15 \times 1}}{{\bf{0}}^{{\bf{ - 6}}}}\)g) of penicillinase, determine the order of the reaction with respect to penicillin and the value of the rate constant.

(Penicillin) (M)

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\({\bf{2}}{\bf{.0 \times 1}}{{\bf{0}}^{{\bf{ - 6}}}}\) \(\)

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\({\bf{3}}{\bf{.0 \times 1}}{{\bf{0}}^{{\bf{ - 6}}}}\)

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