Question

Acetaldehyde decomposes when heated to yield methane and carbon monoxide according to the equation: \({\bf{C}}{{\bf{H}}_{\bf{3}}}{\bf{CHO}}\)(g) ⟶\({\bf{C}}{{\bf{H}}_{\bf{4}}}\)(g) +\({\bf{CO}}\)(g)

Determine the rate law and the rate constant for the reaction from the following experimental data:

Trial	(\({\bf{C}}{{\bf{H}}_{\bf{3}}}{\bf{CHO}}\)) (mol/L)	\(\frac{{ - \Delta \left( {{\bf{C}}{{\bf{H}}_{\bf{3}}}{\bf{CHO}}} \right)}}{{\Delta t}}\)(mol )(Ls−1)
1.	1.75 × 10−3	2.06 × 10−11
2.	3.50 × 10−3	8.24 × 10−11
3.	7.00 × 10−3	3.30 × 10−10

Short answer

The rate\(r = k{\left( {{\bf{C}}{{\bf{H}}_{\bf{3}}}{\bf{CHO}}} \right)^2}\) with rate constant k = 6.73 × 10⁻⁶ L/mol/s.

Step-by-step solution

Rate of a Reaction

The rate of reaction may be defined as the speed of the reactant reacting to obtain a product in a particular reaction at a particular time. The concentration of the reactant and product are represented interms of mole/L.

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

Value of r \(\)

Determine the value of m from the data in which (\({\bf{C}}{{\bf{H}}_{\bf{3}}}{\bf{CHO}}\)) varies. In the last three experiments, (\({\bf{C}}{{\bf{H}}_{\bf{3}}}{\bf{CHO}}\)) varies. When (\({\bf{C}}{{\bf{H}}_{\bf{3}}}{\bf{CHO}}\)) doubles from trial 1.75 to 3.5, the rate doubles, and when \({\bf{C}}{{\bf{H}}_{\bf{3}}}{\bf{CHO}}\) doubles from trial 3.5 to 7, the rate also triples. Thus, the rate is also directly proportional to (\({\bf{C}}{{\bf{H}}_{\bf{3}}}{\bf{CHO}}\)), and m in the rate law is equal to 2.

\(r = k{\left( {{\bf{C}}{{\bf{H}}_{\bf{3}}}{\bf{CHO}}} \right)^2}\)

Value of k

Rate, \(r = k{\left( {{\bf{C}}{{\bf{H}}_{\bf{3}}}{\bf{CHO}}} \right)^2}\)

\(\begin{aligned}{}{\bf{k = }}\frac{{\bf{r}}}{{{{\left( {{\bf{C}}{{\bf{H}}_{\bf{3}}}{\bf{CHO}}} \right)}^{\bf{2}}}}}{\bf{ }}\\\begin{aligned}{{}{}}{{\bf{k = }}\frac{{{\bf{2}}{\bf{.06 \times 1}}{{\bf{0}}^{{\bf{ - 11}}}}\left( {{\bf{mol }}} \right)\left( {{\bf{L}}{{\bf{s}}^{{\bf{ - 1}}}}} \right)}}{{{{({\bf{1}}{\bf{.75 \times 1}}{{\bf{0}}^{{\bf{ - 3}}}})}^{\bf{2}}}{\bf{mol/L}}}}{\bf{ }}}\\{{\bf{k = 6}}{\bf{.73 \times 1}}{{\bf{0}}^{{\bf{ - 6}}}}{\bf{L/mol/s}}{\bf{.}}}\end{aligned}\end{aligned}\)