The rate constant for the rate of decomposition of \({{\bf{N}}_{\bf{2}}}{{\bf{O}}_{\bf{5}}}\)to\({\bf{NO}}\) and \({{\bf{O}}_{\bf{2}}}\)in the gas phase is 1.66 L/mol/s at 650 K and 7.39 L/mol/s at 700 K:

\({\bf{2}}{{\bf{N}}_{\bf{2}}}{{\bf{O}}_{\bf{5}}}{\bf{(g) - - - 4NO(g) + 3}}{{\bf{O}}_{\bf{2}}}{\bf{(g)}}\)

Assuming the kinetics of this reaction are consistent with the Arrhenius equation, calculate the activation energy for this decomposition.

Short Answer

Expert verified

The activation energy for this decomposition is 113,000 J/mol.

Step by step solution

01

Rate of a Reaction

The rate of a reaction can be obtained from the stoichiometry of the reaction.

It is expressed in terms of the change in the amount of any reactant or product, and may be simply derived

\({\bf{rate = k}}{\left( {\bf{A}} \right)^{\bf{m}}}{\left( {\bf{B}} \right)^{\bf{n}}}^{}\)

02

Activation Energy            \(\)

The minimum amount of energy (or threshold energy) needed to activate or energize molecules or atoms to undergo a chemical reaction or transformation.

\(\)

03

Explanation

Rate constant:

\({{\bf{k}}_{\bf{1}}}\)= 1.66 L/mol/s at Temperature,\({{\bf{T}}_1}\)= 650 K

\({{\bf{k}}_2}\)= 7.39 L/mol/s at Temperature, \({{\bf{T}}_{\bf{2}}}\) = 700 K

Taking Logarithm of rate constants,

\(\begin{aligned}{{}{}}{{\bf{ln}}\left( {{\bf{1}}{\bf{.66}}} \right){\bf{ = 0}}{\bf{.5068}}}\\{{\bf{ln}}\left( {{\bf{7}}{\bf{.39}}} \right){\bf{ = 2}}{\bf{.0001}}}\\{{\bf{ln }}{{\bf{K}}_{\bf{2}}}{\bf{-- ln }}{{\bf{K}}_{\bf{1}}}{\bf{ = 2}}{\bf{.0001 -- 0}}{\bf{.5068 = 1}}{\bf{.4933}}}\end{aligned}\)

Temperature:

\(\begin{aligned}{}\begin{aligned}{{}{}}{\left( {\frac{{\bf{1}}}{{{{\bf{T}}_{\bf{2}}}}}} \right){\bf{ -- }}\left( {\frac{{\bf{1}}}{{{{\bf{T}}_{\bf{1}}}}}} \right){\bf{ = }}\left( {\frac{{\bf{1}}}{{{\bf{700}}}}} \right){\bf{ -- }}\left( {\frac{{\bf{1}}}{{{\bf{650}}}}} \right)}\\{\left( {\frac{{\bf{1}}}{{{{\bf{T}}_{\bf{2}}}}}} \right){\bf{ -- }}\left( {\frac{{\bf{1}}}{{{{\bf{T}}_{\bf{1}}}}}} \right){\bf{ = 0}}{\bf{.00143 -- 0}}{\bf{.00154}}}\end{aligned}\\\left( {\frac{{\bf{1}}}{{{{\bf{T}}_{\bf{2}}}}}} \right){\bf{ -- }}\left( {\frac{{\bf{1}}}{{{{\bf{T}}_{\bf{1}}}}}} \right){\bf{ = - 0}}{\bf{.0011}}\end{aligned}\)

Gas Constant, R= −8.314 J\({\bf{mo}}{{\bf{l}}^{{\bf{ - 1}}}}{{\bf{K}}^{{\bf{ - 1}}}}\)

Activation Energy,

\(\begin{aligned}{{}{}}{{{\bf{E}}_{\bf{a}}}{\bf{ = - 8}}{\bf{.314J mo}}{{\bf{l}}^{{\bf{ - 1}}}}{{\bf{K}}^{{\bf{ - 1}}}}{\bf{ \times }}\frac{{\left( {{\bf{ln }}{{\bf{K}}_{\bf{2}}}{\bf{-- ln }}{{\bf{K}}_{\bf{1}}}} \right)}}{{{\bf{ }}\left( {{\bf{1/}}{{\bf{T}}_{\bf{2}}}} \right){\bf{ -- }}\left( {{\bf{1/}}{{\bf{T}}_{\bf{1}}}} \right)}}{\bf{ }}}\\{{{\bf{E}}_{\bf{a}}}{\bf{ = - 8}}{\bf{.314 \times }}\frac{{{\bf{1}}{\bf{.4933 }}}}{{{\bf{0}}{\bf{.00011}}}}}\\\begin{aligned}{}{{\bf{E}}_{\bf{a}}}{\bf{ = 112866}}{\bf{.329 J/mole }}\\\,\,\,\,\,\,\,{\bf{ = 113,000 J/mole}}\end{aligned}\end{aligned}\)

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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 °C:

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

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.

The rate constant for the radioactive decay of 14C is \({\bf{1}}{\bf{.21 \times 1}}{{\bf{0}}^{{\bf{ - 4}}}}{\bf{ yea}}{{\bf{r}}^{{\bf{ - 1}}}}\). The products of the decay are nitrogen atoms and electrons (beta particles): \(\begin{aligned}{}_{\bf{6}}^{{\bf{14}}}{\bf{C}} \to _{\bf{6}}^{{\bf{14}}}{\bf{N + }}{{\bf{e}}^{\bf{ - }}}\\{\bf{rate = k(}}_{\bf{6}}^{{\bf{14}}}{\bf{C)}}\end{aligned}\).

What is the instantaneous rate of production of N atoms in a sample with a carbon-14 content of \({\bf{ 6}}{\bf{.5 \times 1}}{{\bf{0}}^{{\bf{ - 9 }}}}{\bf{M}}\)?

How will each of the following affect the rate of the reaction:

\({\bf{CO}}\left( {\bf{g}} \right){\bf{ + \;N}}{{\bf{O}}_{\bf{2}}}\left( {\bf{g}} \right) \to {{\bf{O}}_{\bf{2}}}{\bf{\;}}\left( {\bf{g}} \right){\bf{ + NO}}\left( {\bf{g}} \right)\) if the rate law for the reaction is rate = \({\bf{k(NO}}{}_{\bf{2}}{\bf{)(CO)}}\)?

  1. Increasing the pressure of \({\bf{NO}}{}_{\bf{2}}\) from 0.1 atm to 0.3 atm
  2. Increasing the concentration of CO from 0.02 M to 0.06 M.

In the PhET Reactions & Rates (http://openstaxcollege.org/l/16PHETreaction) interactive, on the Many Collisions tab, set up a simulation with 15 molecules of A and 10 molecules of BC. Select “Show Bonds” under Options.

  1. Leave the Initial Temperature at the default setting. Observe the reaction. Is the rate of reaction fast or slow?
  2. Click “Pause” and then “Reset All,” and then enter 15 molecules of A and 10 molecules of BC once again. Select “Show Bonds” under Options. This time, increase the initial temperature until, on the graph, the total average energy line is completely above the potential energy curve. Describe what happens to the reaction
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