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From the following data, determine the rate law, the rate constant, and the order with respect to A for the reaction \({\bf{A}} \to {\bf{2C}}\).

Short Answer

Expert verified

The rate law for the given reaction is represented as the rate of reaction =\({\bf{k(A}}{{\bf{)}}^{\bf{2}}}\)

The overall order of the reaction is 2.

The value of the rate constant is \({\bf{2}}{\bf{.14 \times 1}}{{\bf{0}}^{{\bf{ - 3}}}}{\bf{mol}}{{\bf{L}}^{{\bf{ - 1}}}}{{\bf{h}}^{{\bf{ - 1}}}}\)

Step by step solution

01

Rate law

Rate law: General rate of reaction = \({\bf{k(A}}{{\bf{)}}^{\bf{m}}}\)

\(\begin{align}Experiment\,1:3.80 \times {10^{ - 7}}mol{L^{ - 1}}{h^{ - 1}} &= k{(1.33 \times {10^{ - 2}})^m}\\Experiment\,2:1.52 \times {10^{ - 6}}mol{L^{ - 1}}{h^{ - 1}} &= k{(2.66 \times {10^{ - 2}})^m}\\Experiment\,3:3.42 \times {10^{ - 6}}mol{L^{ - 1}}{h^{ - 1}} &= k{(3.99 \times {10^{ - 2}})^m}\end{align}\)

Value of m examining from experiments 1 and 2. It found the rate increased by a factor of 4 and concentration is increased by a factor of 2

Value of m examining from experiments 1 and 3. It found the rate increased by a factor of 9, and concentration is increased by a factor of 3.

02

Order of reaction (m)

To calculate the overall order of reaction, equation (2) is divided by (1), we get

\(\begin{align}\frac{{3.80 \times {{10}^{ - 7}}mol{L^{ - 1}}{h^{ - 1}} = k{{(1.33 \times {{10}^{ - 2}})}^m}}}{{1.52 \times {{10}^{ - 6}}mol{L^{ - 1}}{h^{ - 1}} = k{{(2.66 \times {{10}^{ - 2}})}^m}}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{1}{4} &= {(\frac{1}{2})^m}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,m &= 2\end{align}\)

Since the value of m is 2 hence, the order of the reaction will be second.

Reactions in which reactants are identical and form a product can also be a second-order reaction.

A second-order reaction rate is proportional to the square of the concentration of a reactant.

Hence, the rate of reaction for a given reaction is

Rate of reaction =\({\bf{k(A}}{{\bf{)}}^{\bf{2}}}\)

03

Rate constant (k)

The rate constant is the proportionality constant in the equation that expresses the relationship between the rate of a chemical reaction and the concentration of reacting substances.

From the given table, the value of the rate constant can be calculated as;

For experiment 1, rate constant

\(k = \frac{{3.80 \times {{10}^{ - 7}}mol{L^{ - 1}}{h^{ - 1}}}}{{{{(1.33 \times {{10}^{ - 2}})}^2}}} = 2.14 \times {10^{ - 3}}mol{L^{ - 1}}{h^{ - 1}}\)

For experiment 2, rate constant

\(k = \frac{{1.52 \times {{10}^{ - 6}}mol{L^{ - 1}}{h^{ - 1}}}}{{{{(2.66 \times {{10}^{ - 2}})}^2}}} = 2.14 \times {10^{ - 3}}mol{L^{ - 1}}{h^{ - 1}}\)

For experiment 3, rate constant

\(k = \frac{{3.42 \times {{10}^{ - 6}}mol{L^{ - 1}}{h^{ - 1}}}}{{{{(3.99 \times {{10}^{ - 2}})}^2}}} = 2.14 \times {10^{ - 3}}mol{L^{ - 1}}{h^{ - 1}}\)

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Most popular questions from this chapter

For the reaction\({\bf{A}} \to {\bf{B + C}}\), the following data were obtained at 30 ยฐC:

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

The half-life of a reaction of compound A to give compounds D and E is 8.50 min when the initial concentration of A is 0.150 mol/L. How long will it take for the concentration to drop to 0.0300 mol/L if the reaction is (a) first order with respect to A or (b) second order with respect to A?

The decomposition of acetaldehyde is a second-order reaction with a rate constant of \({\bf{4}}{\bf{.71 \times 1}}{{\bf{0}}^{{\bf{ - 8 }}}}{\bf{L mo}}{{\bf{l}}^{{\bf{ - 1}}}}{\bf{ s}}{{\bf{ }}^{{\bf{ - 1}}}}\). What is the instantaneous rate of decomposition of acetaldehyde in a solution with a concentration of \({\bf{5}}{\bf{.55 \times 1}}{{\bf{0}}^{{\bf{ - 4}}}}{\bf{M}}\)?

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.

Describe the effect of each of the following on the rate of the reaction of magnesium metal with a solution of hydrochloric acid: the molarity of the hydrochloric acid, the temperature of the solution, and the size of the pieces of magnesium.

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