Question

The rate constant for the decomposition of acetaldehyde, \({\bf{C}}{{\bf{H}}_{\bf{3}}}{\bf{CHO}}\), to methane, \({\bf{C}}{{\bf{H}}_{\bf{4}}}\), and carbon monoxide, CO, in the gas phase is 1.1 × 10⁻² L/mol/s at 703 K and 4.95 L/mol/s at 865 K. Determine the activation energy for this decomposition.

Short answer

The energy of activation for the reaction is 96.769 kJ/mol

Step-by-step solution

Step 1: Using Arrhenius Equation

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

\(\log \frac{{{k_2}}}{{{k_1}}} = \frac{{{E_a}}}{{2.303*R}}\left( {\frac{1}{{{T_1}}} - \frac{1}{{{T_2}}}} \right)\)

where k₁and k₂are 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

Calculation

Replacing the values,

\(\log \frac{{4.95}}{{1.1*{{10}^{ - 2}}}} = \frac{{{E_a}}}{{2.303*8.314}}\left( {\frac{1}{{703}} - \frac{1}{{865}}} \right)\)

\({E_a} = 190691.9J/mol\)

\({E_a} = 190.69kJ/mol\)