Question

The decomposition of many substances on the surface of a heterogeneous catalyst shows the following behaviour:

How do you account for the rate law changing from first order to zero order in the concentration of reactant?

a. Determine the first-order rate constants for thermal degradation of silk for each of the three experiments.

b. Does either of the two additives appear to retard the degradation of silk?

c. Calculate the half-life for the thermal degradation of silk for each of the three experiments.

Short answer

Heterogeneous catalysisby solids plays a key role in chemical industry due to the ability to accelerate a chemical reaction to reach the equilibrium and to selectively accelerate a desirable chemical reaction in a complex chemical reaction network to reach the equilibrium.

a) untreated$K=0.465da{y}^{-1}$

deacidifying $K=0.659da{y}^{-1}$

antioxidant$K=0.779da{y}^{-1}$

b) Both additives are increasing the rate of degradation process

c) untreated$t_{1/2}=\frac{\ln 2}{0.465days}=1.49days$

deacidifying $t_{1/2}=\frac{\ln 2}{0.659day{s}^{-1}}=1.05days$

antioxidant$t_{1/2}=\frac{\ln 2}{0.779day{s}^{-1}}=0.890days$

Step-by-step solution

Determine the reaction between the reactant and the catalyst

The reaction happens on the surface of the catalyst followed by desorption of the product. The rate of reaction is represented as follows:

$Rate=\frac{k_1{k}_2{C}_A}{1+k_1{C}_A}$

Here,$C_A$ is the concentration of reactant A and $C_s$is the active site on the catalyst S.

The reaction is represented as follows:

$Rate=\frac{k_1{k}_2{C}_A}{K_1{C}_A}$

$Rate=\frac{k_1{k}_2}{k_1}$

Here, no reactant molecule is involved in the rate expression; therefore, the reaction is zero-order in this case.

Determine the zero-order reaction.

The integrated rate law for the zero-order reaction A → products is [A]_t = -kt + [A]. Because this equation has the form y = mx + b, a plot of the concentration of A as a function of time yields a straight line.

The rate constant for the reaction can be determined from the slope of the line, which is equal to -k.