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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 \({\bf{5}}{\bf{.55 \times 1}}{{\bf{0}}^{{\bf{ - 4}}}}{\bf{M}}\)?

Short Answer

Expert verified

The instantaneous rate of decomposition of aldehyde is \({\bf{1}}{\bf{.45}}\,{\bf{ \times 1}}{{\bf{0}}^{{\bf{ - 13}}}}\,{\bf{mol}}{{\bf{L}}^{{\bf{ - 1}}}}{{\bf{s}}^{{\bf{ - 1}}}}\).

Step by step solution

01

Definitions

The rate law for a chemical reaction is an expression that provides a relationship between the rate of the reaction and the concentration of the reactants participating in it. It can also be a second-order reaction.

An instantaneous rate of change is the change in the rate at a particular instant.

02

Rate of reaction

The decomposition of aldehyde is a second-order reaction.

The rate law of reaction can be represented as

\({\bf{rate}}\;{\bf{of}}\;{\bf{reaction = k(acetaldehyde}}{{\bf{)}}^{\bf{2}}}\)

Given the value of the rate constant is\({\bf{4}}{\bf{.71 \times 1}}{{\bf{0}}^{{\bf{ - 8 }}}}{\bf{L mo}}{{\bf{l}}^{{\bf{ - 1}}}}{\bf{ s}}{{\bf{ }}^{{\bf{ - 1}}}}\)and the concentration of aldehyde \({\bf{5}}{\bf{.55 \times 1}}{{\bf{0}}^{{\bf{ - 4}}}}{\bf{M}}\).

The instantaneous rate of decomposition is equal to the rate of reaction.

Hence the rate of reaction can be calculated as follow

\(\begin{aligned}{}{\bf{Rate}}\,\,{\bf{ = k(Acetaldehyde}}{{\bf{)}}^{\bf{2}}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,{\bf{ = 4}}{\bf{.71 \times 1}}{{\bf{0}}^{{\bf{ - 7}}}}{\bf{Lmo}}{{\bf{l}}^{{\bf{ - 1}}}}{{\bf{s}}^{{\bf{ - 1}}}}{{\bf{(5}}{\bf{.55 \times 1}}{{\bf{0}}^{{\bf{ - 4}}}}{\bf{mol}}{{\bf{L}}^{{\bf{ - 1}}}}{\bf{)}}^{\bf{2}}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,{\bf{ = 1}}{\bf{.45}}\,{\bf{ \times 1}}{{\bf{0}}^{{\bf{ - 13}}}}\,{\bf{mol}}{{\bf{L}}^{{\bf{ - 1}}}}{{\bf{s}}^{{\bf{ - 1}}}}\end{aligned}\)

Thus, the instantaneous rate of decomposition is \({\bf{1}}{\bf{.45}}\,{\bf{ \times 1}}{{\bf{0}}^{{\bf{ - 13}}}}\,{\bf{mol}}{{\bf{L}}^{{\bf{ - 1}}}}{{\bf{s}}^{{\bf{ - 1}}}}\).

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Most popular questions from this chapter

Acetaldehyde decomposes when heated to yield methane and carbon monoxide according to the equation: \({\bf{C}}{{\bf{H}}_{\bf{3}}}{\bf{CHO}}\)(g) ⟶\({\bf{C}}{{\bf{H}}_{\bf{4}}}\)(g) +\({\bf{CO}}\)(g)

Determine the rate law and the rate constant for the reaction from the following experimental data:

Trial

(\({\bf{C}}{{\bf{H}}_{\bf{3}}}{\bf{CHO}}\)) (mol/L)

\(\frac{{ - \Delta \left( {{\bf{C}}{{\bf{H}}_{\bf{3}}}{\bf{CHO}}} \right)}}{{\Delta t}}\)(mol )(Ls−1)

1.

1.75 × 10−3

2.06 × 10−11

2.

3.50 × 10−3

8.24 × 10−11

3.

7.00 × 10−3

3.30 × 10−10

Use the provided initial rate data to derive the rate law for the reaction whose equation is: \({\bf{OC}}{{\bf{l}}^ - }\)(aq) + \({{\bf{I}}^ - }\)(aq) ⟶OI−(aq) +\({\bf{C}}{{\bf{l}}^ - }\)(aq)

Trial

(\({\bf{OC}}{{\bf{l}}^ - }\)) (mol/L)

(\({{\bf{I}}^ - }\)) (mol/L)

Initial Rate (mol/L/s)

1.

0.0040

0.0020

0.00184

2.

0.0020

0.0040

0.00092

3.

0.0020

0.0020

0.00046

Determine the rate law expression and the value of the rate constant k with appropriate units for this reaction.

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.

The reaction of compound A to give compounds C and D was found to be second-order in A. The rate constant for the reaction was determined to be 2.42 L/mol/s. If the initial concentration is 0.500 mol/L, what is the value of t1/2?

Atomic chlorine in the atmosphere reacts with ozone in the following pair of elementary reactions:

\({\bf{Cl + \;O3}}\left( {\bf{g}} \right){\bf{ - - - ClO}}\left( {\bf{g}} \right){\bf{ + \;O2}}\left( {\bf{g}} \right)\)(rate constant k1 )

\({\bf{ClO}}\left( {\bf{g}} \right){\bf{ + O - - - Cl}}\left( {\bf{g}} \right){\bf{ + \;O2}}\left( {\bf{g}} \right)\)(rate constant k2 )

Determine the overall reaction, write the rate law expression for each elementary reaction, identify any intermediates, and determine the overall rate law expression.

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