/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q49 E Nitro-glycerine is an extremely ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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

Expert verified

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

01

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.

02

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:

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

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

When every collision between reactants leads to a reaction, what determines the rate at which the reaction occurs?

What is the activation energy of a reaction, and how is this energy related to the activated complex of the reaction?

Consider the following reaction in aqueous solution:

\(\)

\({\bf{5B}}{{\bf{r}}^ - }\left( {aq} \right) + BrO_3^ - \left( {aq} \right) + 6{H^ + }\left( {aq} \right) \to 3B{r_2}\left( {aq} \right) + 3{H_2}O\left( l \right)\)

If the rate of disappearance of Br– (aq) at a particular moment during the reaction is 3.5 × 10−4 M s −1, what is the rate of appearance of\({\bf{B}}{{\bf{r}}_{\bf{2}}}\)(aq) at that moment?

For the past 10 years, the unsaturated hydrocarbon 1,3-butadiene \(\left( {{\bf{C}}{{\bf{H}}_{\bf{2}}}{\bf{ = CH - CH = C}}{{\bf{H}}_{\bf{2}}}} \right)\) has ranked 38th among the top 50 industrial chemicals. It is used primarily for the manufacture of synthetic rubber. An isomer exists also as cyclobutene:

The isomerization of cyclobutene to butadiene is first-order, and the rate constant has been measured as \({\bf{2}}{\bf{.0 \times 1}}{{\bf{0}}^{{\bf{ - 4}}}}{{\bf{s}}^{{\bf{ - 1}}}}\) at 150 \({\bf{^\circ C}}\) in a 0.53-L flask. Determine the partial pressure of cyclobutene and its concentration after 30.0 minutes if an isomerization reaction is carried out at 150 \({\bf{^\circ C}}\) with an initial pressure of 55 torr.

If the rate of decomposition of ammonia, \({\bf{N}}{{\bf{H}}_{\bf{3}}}\), at 1150 K is \(2.10 \times 1{0^{ - 6}}mol/L/s\), what is the rate of production of nitrogen and hydrogen?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.