91影视

From the Arrhenius Equation, the rate of reaction at two different temperatures is given as

\({\bf{log}}\frac{{{{\bf{k}}_{\bf{2}}}}}{{{{\bf{k}}_{\bf{1}}}}}{\bf{ = }}\frac{{{{\bf{E}}_{\bf{a}}}}}{{{\bf{2}}{\bf{.303*R}}}}\left( {\frac{{\bf{1}}}{{{{\bf{T}}_{\bf{1}}}}}{\bf{ - }}\frac{{\bf{1}}}{{{{\bf{T}}_{\bf{2}}}}}} \right)\)

where \({{\bf{k}}_{\bf{1}}}\) and \({{\bf{k}}_{\bf{2}}}\) are the rate constants at \({{\bf{T}}_{\bf{1}}}\) and \({{\bf{T}}_{\bf{2}}}\) where \({{\bf{T}}_{\bf{1}}}{\bf{ < }}{{\bf{T}}_{\bf{2}}}\). \({{\bf{E}}_{\bf{a}}}\)is the activation energy (in J) and R is the gas constant.

From the question, the ratio of \(\frac{{{{\bf{k}}_{\bf{2}}}}}{{{{\bf{k}}_{\bf{1}}}}}{\bf{ = 1}}{\bf{.47}}\)

Replacing the values in the Arrhenius equation,

\(\begin{align}\log \frac{{8.5*{{10}^{ - 11}}}}{{2.1*{{10}^{ - 11}}}} &= \frac{{{E_a}}}{{2.303*8.314}}\left( {\frac{1}{{300}} - \frac{1}{{310}}} \right)\\{E_a} &= 0.607*2.303*8.314*\left( {\frac{{300*310}}{{310 - 300}}} \right)\\{E_a} &= 108123J/mol\\{E_a} &= 108.12kJ/mol\end{align}\)

The Arrhenius equation is given as \({\bf{k = A}}{e^{\frac{{{\bf{ - }}{{\bf{E}}_{\bf{a}}}}}{{{\bf{RT}}}}}}\) ,where A is the frequency factor, k is the rate constant, \({{\bf{E}}_{\bf{a}}}\) is the activation energy, R is the gas constant, and T is the temperature in Kelvin.

The frequency factor can be calculated by considering either rate constant value given. Considering the rate constant at 37^oC and replacing the values in the equation.

\({\bf{8}}{\bf{.5*1}}{{\bf{0}}^{{\bf{ - 11}}}}{\bf{ = A}}{{\bf{e}}^{\frac{{{\bf{ - 108123}}}}{{{\bf{8}}{\bf{.314*310}}}}}}\)

\( \Rightarrow {\bf{8}}{\bf{.5*1}}{{\bf{0}}^{{\bf{ - 11}}}}{\bf{ = A}}{{\bf{e}}^{{\bf{ - 41}}{\bf{.95}}}}\)

\( \Rightarrow {\bf{A = }}\frac{{{\bf{8}}{\bf{.5*1}}{{\bf{0}}^{{\bf{ - 11}}}}}}{{{\bf{1}}{\bf{.65*1}}{{\bf{0}}^{{\bf{ - 19}}}}}}\)

\( \Rightarrow {\bf{A = 1}}{\bf{.408*1}}{{\bf{0}}^{\bf{8}}}{{\bf{s}}^{{\bf{ - 1}}}}\)

Again, from the Arrhenius equation

\({\bf{k = A}}{e^{\frac{{{\bf{ - }}{{\bf{E}}_{\bf{a}}}}}{{{\bf{RT}}}}}}\)

where A is the frequency factor (calculated previously), k is the rate constant, \({{\bf{E}}_{\bf{a}}}\) is the activation energy, R is the gas constant, and T is the temperature in Kelvin.

At 47^oC, the rate can be calculated as

\(k = 1.408*{10^8}*{e^{\frac{{ - 108123}}{{8.314*320}}}}\)

\(\begin{align}\Rightarrow k &= 1.408*{10^8}*{e^{ - 40.6}}\\ \Rightarrow k &= 3.15*{10^{ - 10}}\end{align}\)

The integral rate equation for the first-order reaction is given as follows.

\({\bf{k = }}\frac{{{\bf{2}}{\bf{.303}}}}{{\bf{t}}}{\bf{log}}\frac{{{{\left( {\bf{A}} \right)}_{\bf{o}}}}}{{\left( {\bf{A}} \right)}}\)

Where (A)₀ is the initial concentration and (A) is the concentration at time t.

Here, \({\left( {\bf{A}} \right)_{\bf{0}}}{\bf{ = 0}}{\bf{.150M}}\) and \(\left( {\bf{A}} \right){\bf{ = 1}}{\bf{.65 \times 1}}{{\bf{0}}^{{\bf{ - 7}}}}\)^.Replacing the values in the above equation,

At k=\({\bf{2}}{\bf{.1 \times 1}}{{\bf{0}}^{{\bf{ - 11}}}}\) at T=27⁰C

\(\begin{align}\Rightarrow t &= \frac{{2.303}}{k}\log \frac{{{{\left( A \right)}_o}}}{{\left( A \right)}}\\ \Rightarrow t &= \frac{{2.303}}{{2.1*{{10}^{ - 11}}}}\log \frac{{0.150}}{{1.65*{{10}^{ - 7}}}}\\ \Rightarrow t &= 6.53*{10^{11}}{s^{}}\\ \Rightarrow t &= \frac{{6.53*{{10}^{11}}}}{{60*60*24}}days\\ \Rightarrow t &= 7.56*{10^6}days\end{align}\)

The assumption that the reaction is irreversible simplifies the calculation as we do not have to worry about the product being converted back to the reactants.

The assumption that the reaction is irreversible simplifies the calculation as we do not have to worry about the product being converted back to the reactants.

Since equilibrium is not set up in the irreversible reaction, the rate law expression can be used directly without any modification.