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Chapter 12: Q39 E (page 706)

What is the half-life for the decomposition of NOCl when the concentration of NOCl is 0.15 M? The rate constant for this second-order reaction is \({\bf{8}}{\bf{.0 \times 1}}{{\bf{0}}^{{\bf{ - 8}}}}\)L/mol/s.

Short Answer

Expert verified

Half-life periods may be defined as the time period at which half the concentration of reactant converts into the product. The half-life period of the second-order decomposition of the NOCl is 93750000 years which has a rate constant \({\bf{8}}{\bf{.0 \times 1}}{{\bf{0}}^{{\bf{ - 8}}}}\)L/mol/s.

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Step by step solution

01

Reaction Rate

The reaction involved the effective collision of two reactants to produce the desired products. Reactions can be natural, which occur in the surrounding environment, whereas it can be artificially done in the laboratory to form a desired required product.

The reaction rate can be defined as the speed of reaction to produce the products. The reaction rate can be slow, fast or moderate. The reaction can take less than a millisecond to produce products, or it can take years to produce the desired product.

The half-life period can be defined as the time period at which half the concentration of the reactants gets converted into a product.

02

Explanation

The half-life period of the second-order decomposition of the NOCl is:

\({\bf{Half - life period = }}\frac{{\left( {\bf{A}} \right){\bf{^\circ }}}}{{{\bf{2k}}}}\)

Where \(\left( {\bf{A}} \right){\bf{^\circ }}\) = initial concentration of reactant

K = Rate constant

Initial concentration of reactant,\(\left( {\bf{A}} \right){\bf{^\circ }}\) =0.15 M

Rate constant for the decomposition of NOCl =\({\bf{8}}{\bf{.0 \times 1}}{{\bf{0}}^{{\bf{ - 8}}}}\) L/mol/s

\({\bf{Half - life period = }}\frac{{{\bf{0}}{\bf{.15}}}}{{{\bf{2 \times }}\left( {{\bf{8}}{\bf{.0 \times 1}}{{\bf{0}}^{{\bf{ - 8}}}}{\bf{yea}}{{\bf{r}}^{{\bf{ - 1}}}}} \right)}}\)

Half-life period = 93750000 years.

Therefore, the half-life period of the second-order decomposition of the NOCl is 93750000 years which has a rate constant is \({\bf{8}}{\bf{.0 \times 1}}{{\bf{0}}^{{\bf{ - 8}}}}\)L/mol/s.

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Most popular questions from this chapter

For the reaction\({\bf{Q}} \to {\bf{W + X}}\), the following data were obtained at 30 °C

  1. What is the order of the reaction with respect to (Q), and what is the rate law?
  2. What is the rate constant?

Experiments were conducted to study the rate of the reaction represented by this equation.(2)\({\rm{2NO(g) + 2}}{{\rm{H}}_{\rm{2}}}{\rm{(g) }} \to {\rm{ }}{{\rm{N}}_{\rm{2}}}{\rm{(g) + 2}}{{\rm{H}}_{\rm{2}}}{\rm{O(g)}}\)Initial concentrations and rates of reaction are given here.

Experiment Initial Concentration

\(\left( {{\bf{NO}}} \right){\rm{ }}\left( {{\bf{mol}}/{\bf{L}}} \right)\)

Initial Concentration, \(\left( {{{\bf{H}}_{\bf{2}}}} \right){\rm{ }}\left( {{\bf{mol}}/{\bf{L}}} \right)\)Initial Rate of Formation of \({{\bf{N}}_{\bf{2}}}{\rm{ }}\left( {{\bf{mol}}/{\bf{L}}{\rm{ }}{\bf{min}}} \right)\)
1\({\bf{0}}.{\bf{0060}}\)\({\bf{0}}.{\bf{00}}1{\bf{0}}\)\({\bf{1}}.{\bf{8}} \times {\bf{1}}{{\bf{0}}^{ - {\bf{4}}}}\)
2\({\bf{0}}.{\bf{0060}}\)\({\bf{0}}.{\bf{00}}2{\bf{0}}\)\({\bf{3}}.{\bf{6}} \times {\bf{1}}{{\bf{0}}^{ - {\bf{4}}}}\)
3\({\bf{0}}.{\bf{00}}1{\bf{0}}\)\({\bf{0}}.{\bf{0060}}\)\({\bf{0}}.{\bf{30}} \times {\bf{1}}{{\bf{0}}^{ - {\bf{4}}}}\)
4\({\bf{0}}.{\bf{00}}2{\bf{0}}\)\({\bf{0}}.{\bf{0060}}\)\({\bf{1}}.{\bf{2}} \times {\bf{1}}{{\bf{0}}^{ - {\bf{4}}}}\)

Consider the following questions:(a) Determine the order for each of the reactants, \({\bf{NO}}\) and \({{\bf{H}}_{\bf{2}}}\), from the data given and show your reasoning.(b) Write the overall rate law for the reaction.(c) Calculate the value of the rate constant, k, for the reaction. Include units.(d) For experiment 2, calculate the concentration of \({\bf{NO}}\)remaining when exactly one-half of the original amount of \({{\bf{H}}_{\bf{2}}}\) had been consumed.(e) The following sequence of elementary steps is a proposed mechanism for the reaction.Step 1:Step 2:Step 3:Based on the data presented, which of these is the rate determining step? Show that the mechanism is consistent with the observed rate law for the reaction and the overall stoichiometry of the reaction.

In a transesterification reaction, a triglyceride reacts with an alcohol to form an ester and glycerol. Many students learn about the reaction between methanol (\({\bf{C}}{{\bf{H}}_{\bf{3}}}{\bf{OH}}\)) and ethyl acetate (\({\bf{C}}{{\bf{H}}_{\bf{3}}}{\bf{C}}{{\bf{H}}_{\bf{2}}}{\bf{OCOC}}{{\bf{H}}_{\bf{3}}}\)) as a sample reaction before studying the chemical reactions that produce biodiesel:

\({\bf{C}}{{\bf{H}}_{\bf{3}}}{\bf{OH + C}}{{\bf{H}}_{\bf{3}}}{\bf{C}}{{\bf{H}}_{\bf{2}}}{\bf{OCOC}}{{\bf{H}}_{\bf{3}}}{\bf{ - - - C}}{{\bf{H}}_{\bf{3}}}{\bf{OCOC}}{{\bf{H}}_{\bf{3}}}{\bf{ + C}}{{\bf{H}}_{\bf{3}}}{\bf{C}}{{\bf{H}}_{\bf{2}}}{\bf{OH}}\).The rate law for the reaction between methanol and ethyl acetate is, under certain conditions, determined to be: rate =\(k\left( {{\bf{C}}{{\bf{H}}_{\bf{3}}}{\bf{OH }}} \right)\). What is the order of reaction with respect to methanol and ethyl acetate, and what is the overall order of reaction?

Given the following reactions and the corresponding rate laws, in which of the reactions might the elementary reaction and the overall reaction be the same?\(\begin{aligned}{\rm{(a) C}}{{\rm{l}}_2}{\rm{ + CO }} \to {\rm{ C}}{{\rm{l}}_2}{\rm{CO}}\\{\rm{rate = }}k{{\rm{(C}}{{\rm{l}}_2}{\rm{)}}^{\frac{3}{2}}}{\rm{(CO)}}\\{\rm{(b) PC}}{{\rm{l}}_3}{\rm{ + C}}{{\rm{l}}_{\rm{2}}}{\rm{ }} \to {\rm{ PC}}{{\rm{l}}_{\rm{5}}}\\{\rm{rate = }}k{\rm{(PC}}{{\rm{l}}_{\rm{3}}}{\rm{) (C}}{{\rm{l}}_{\rm{2}}}{\rm{)}}\\{\rm{(c) 2NO + }}{{\rm{H}}_{\rm{2}}}{\rm{ }} \to {\rm{ }}{{\rm{N}}_{\rm{2}}}{\rm{ + }}{{\rm{H}}_{\rm{2}}}{\rm{O}}\\{\rm{rate = }}k{\rm{(NO)(}}{{\rm{H}}_{\rm{2}}}{\rm{)}}\\{\rm{(d) 2NO + }}{{\rm{O}}_{\rm{2}}}{\rm{ }} \to {\rm{ 2N}}{{\rm{O}}_{\rm{2}}}\\{\rm{rate = }}k{{\rm{(NO)}}^{\rm{2}}}{\rm{(}}{{\rm{O}}_{\rm{2}}}{\rm{)}}\\{\rm{(e) NO + }}{{\rm{O}}_{\rm{3}}}{\rm{ }} \to {\rm{ N}}{{\rm{O}}_{\rm{2}}}{\rm{ + }}{{\rm{O}}_{\rm{2}}}\\{\rm{rate = }}k{\rm{(NO)(}}{{\rm{O}}_{\rm{3}}}{\rm{)}}\end{aligned}\)

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)

Rate (mole/L/min)

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

\({\bf{1}}{\bf{.0 \times 1}}{{\bf{0}}^{{\bf{ - 10}}}}\)

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

\({\bf{1}}{\bf{.5 \times 1}}{{\bf{0}}^{{\bf{ - 10}}}}\)

\({\bf{4}}{\bf{.0 \times 1}}{{\bf{0}}^{{\bf{ - 6}}}}\)

\({\bf{2}}{\bf{.0 \times 1}}{{\bf{0}}^{{\bf{ - 10}}}}\)
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