Question

Nitro-glycerine is an extremely sensitive explosive. In a series of carefully controlled experiments, samples of the explosive were heated to 160 °C, and their first-order decomposition was studied. Determine the average rate constants for each experiment using the following data:

Initial (\({{\bf{C}}_{\bf{3}}}{{\bf{H}}_{\bf{5}}}{{\bf{N}}_{\bf{3}}}{{\bf{O}}_{\bf{9}}}\)) (M)	4.88	3.52	2.29	1.81	5.33	4.05	2.95	1.72
t(s)	300	300	300	300	180	180	180	180
% Decomposed	52.0	52.9	53.2	53.9	34.6	35.9	36.0	35.4

Short answer

The rate constant increases as the time period decrease because the rate constant is inversely proportional to the time period taken by the reaction. The average rate constant of the experimental data is \({\bf{0}}{\bf{.0065 se}}{{\bf{c}}^{{\bf{ - 1}}}}\).

Step-by-step solution

Reaction Rate

The reaction involved the effective collision of two reactants to produce the desired products. Reactions can be natural, which occur in the surrounding environment, whereas it can be artificially done in the laboratory to form the desired product.

The reaction rate can be defined as the reaction speed to produce the products. The reaction rate can be slow, fast or moderate. The reaction can take less than millisecond to produce products, or it can take years to produce the desired product.

The half-life period can be defined as the time period at which half the concentration of the reactants gets converted into a product.

Explanation

The half-life period of the first order is:

\({\bf{Half - life period = }}\frac{{{\bf{ln }}\left( {\bf{2}} \right)}}{{\bf{k}}}\)

The rate constant of first-order does not depend upon the concentration of the reactant, but it depends upon the time taken by the reaction.

Taking the first reading of the time = 300 second to calculate the rate constant.

\(\begin{align}k &= {\bf{ }}\frac{{2.303}}{t}Log{\bf{ }}\frac{{\left( A \right)}}{{{{\left( A \right)}_0}}}\\k &= {\bf{ }}\frac{{2.303}}{{300s}}Log{\bf{ }}\frac{{4.88}}{{0.48}}\\k &= {\bf{ }}\frac{{2.303}}{{300s}}Log{\bf{ }}10.2\\k &= {\bf{ }}\frac{{2.303}}{{300s}} \times 1.0086\\k &= {\bf{ }}0.001\end{align}\)

The rate constant of the reaction is 0.001 sec^-1.

Now, take the first reading of the time = 180 second to calculate the rate constant.

\(\begin{align}k &= {\bf{ }}\frac{{2.303}}{t}Log{\bf{ }}\frac{{\left( A \right)}}{{{{\left( A \right)}_0}}}\\k &= {\bf{ }}\frac{{2.303}}{{180s}}Log{\bf{ }}\frac{{5.33}}{{0.65}}\\k &= {\bf{ }}\frac{{2.303}}{{180s}}Log{\bf{ }}8.2\\k &= {\bf{ }}\frac{{2.303}}{{180s}} \times 0.914\\k &= {\bf{ }}0.012\end{align}\)

The rate constant of the reaction is 0.012 sec^-1.

The rate constant increases as the time period decrease because the rate constant is inversely proportional to the time period taken by the reaction.

Therefore, the average rate constant of the reaction is:

\(\begin{align}Average\,Rate\,Cons\tan t &= \frac{{0.001 + 0.012}}{2}\\Average\,Rate\,Cons\tan t &= 0.0065{\sec ^{ - 1}}\end{align}\)