Answer to Question #322518 in Physical Chemistry for Vikram

Question

The decomposition of acetaldehyde in gas phase was studied at 729k using the initial concentration of 7.77×10^-3M and 3.64×10^-3M. The half life periods were determined to be 295s and 515s respectively. Determine the order of the reaction.

Answer 1

t_1/2(1) = 295 s;

t_1/2(2) = 515 s;

C₀(1) = 0.00777 M;

C₀(2) = 0.00364 M;

Order №1:

t_1/2 = ln2/k

k₁ = ln2/t_1/2(1) = 0.693/295 = 0.002;

k₂ = ln2/t_1/2(2) = 0.693/515 = 0.001;

Order №2:

t_1/2 = 1/k * 1/C₀;

k₁ = 1/(t_1/2(1) * C₀(1)) = 1/(295 * 0.00777) = 0.4;

k₂ = 1/(t_1/2(2) * C₀(2)) = 1/(515 * 0.00364) = 0.5;

Order №3:

t_1/2 = 1/k * 3/(2 * C₀²);

k₁ = 3/(2 * C₀(1)² * t_1/2(1)) = 3/(2 * 0.00777² * 295) = 84.2;

k₂ = 3/(2 * C₀(2)² * t_1/2(2)) = 3/(2 * 0.00364² * 515) = 219.;

Therefore: for the reaction determine the second order, k₁ = k₂