Q12.11CYL The rate constant for the rate o... [FREE SOLUTION]

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Chapter 12: Q12.11CYL (page 694)

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

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

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

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.

\(\)

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= 鈭�8.314 J\({\bf{mo}}{{\bf{l}}^{{\bf{ - 1}}}}{{\bf{K}}^{{\bf{ - 1}}}}\)

Activation Energy,

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