Question

The thermal degradation of silk was studied by Kuruppillai, Hersh, and Tucker (“Historic Textile and Paper Materials,”ACS Advances in Chemistry Series, No. 212, 1986) by measuring the tensile strength of silk fibres at various time of exposures to elevated temperature. The loss of tensile strength follows first-order kinetics,

$\mathbf{-}\frac{\mathbf{d}\mathbf{s}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathit{k}\mathit{s}$

Where s is the strength to the fibre retained after heating and k is the first-order rate constant. The effects of adding a deacidifying agent and an antioxidant to the silk were studied, and the following data were obtained:

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

a) untreated$\mathrm{K}=0.465{\mathrm{day}}^{-1}$

Deacidifying $\mathrm{K}=0.659{\mathrm{days}}^{-1}$

Antioxidant $\mathrm{K}=0.779{\mathrm{days}}^{-1}$

b) Both additives are increasing the rate of degradation process

c) Untreated $t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{\ln 2}{0.465day{s}^{-1}}=1.49days$

Deacidifying $t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{\ln 2}{0.659day{s}^{-1}}=1.05days$

Antioxidant $t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{\ln 2}{0.779day{s}^{-1}}=0.890days$

Step-by-step solution

Determine the first order reaction

The following graphs represents concentration of reactants versus time for a first-order reaction.

Plotting $\ln \left[A\right]$ with respect to time for a first-order reaction gives a straight line with the slope of the line equal to -K-K.

the half-life for the thermal degradation of silk

The half-life$\left(t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)$ is a timescale on which the initial population is decreased by half of its original value, represented by the following equation.

$\left[A\right]=12{\left[A\right]}_0\left(2.3.5\right)\left(2.3.5\right)\left[A\right]=12{\left[A\right]}_0$

After a period of one half-life,and we can write

${\left[A\right]}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\left[A\right]}_0=12=e^{-K{t}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left(2.3.6\right)}$

Taking logarithms of both sides (remember that$\ln e^x=x$) yields

$\ln 0.5=-Kt\left(2.3.7\right)$

Solving for the half-life, we obtain the simple relation

$t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\ln 2K\approx 0.693K\left(2.3.8\right)$

This indicates that the half-life of a first-order reaction is a constant.