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Chapter 12: Q59 E (page 708)

The rate constant at 325掳C for the decomposition reaction \({{\bf{C}}_{\bf{4}}}{{\bf{H}}_{\bf{8}}} \to {\bf{2}}{{\bf{C}}_{\bf{2}}}{{\bf{H}}_{\bf{4}}}\)is 6.1 脳 10鈭�8 s鈭�1, and the activation energy is 261 kJ per mole of\({{\bf{C}}_{\bf{4}}}{{\bf{H}}_{\bf{8}}}\). Determine the frequency factor for the reaction.

Short Answer

Expert verified

The frequency factor for the reaction is\({\bf{3}}{\bf{.84*1}}{{\bf{0}}^{{\bf{15}}}}{{\bf{s}}^{{\bf{ - 1}}}}\).

Step by step solution

01

Using Arrhenius Equation

The Arrhenius equation is given as\({\bf{k = A}}{{\bf{e}}^{\frac{{{\bf{ - }}{{\bf{E}}_{\bf{a}}}}}{{{\bf{RT}}}}}}\),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.

02

Calculation of Activation energy

Replacing the values in the reaction,

\(6.1*{10^{ - 8}} = A{e^{\frac{{ - 261000}}{{8.314*598}}}}\)

\( \Rightarrow 6.1*{10^{ - 8}} = A{e^{ - 52.5}}\)

\( \Rightarrow A = \frac{{6.1*{{10}^{ - 8}}}}{{1.589*{{10}^{ - 23}}}}\)

\( \Rightarrow A = 3.84*{10^{15}}{s^{ - 1}}\)

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