/*! 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} Q48 E Recently, the skeleton of King R... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Recently, the skeleton of King Richard III was found under a parking lot in England. If tissue samples from the skeleton contain about 93.79% of the carbon-14 expected in living tissue, what year did King Richard III die? The half-life for carbon-14 is 5730 years.

Short Answer

Expert verified

The half-life period of the carbon-14 in living tissues is 5730 years which has rate constant is 0.00012 year-1 and the time-period taken to kill Richard-III die in 218.88 years or 219 years.

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 a 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

Numerical Explanation

The half-life period of the first-order decay of carbon-14 is:

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

The half-life period of the carbon-14 = 5730 years

\(\begin{align}Half - life{\bf{ }}period{\bf{ }} &= {\bf{ }}\frac{{0.693}}{k}\\5730years{\bf{ }} &= {\bf{ }}\frac{{0.693}}{k}\\k{\bf{ }} &= {\bf{ }}\frac{{0.693}}{{5730years}}\\k &= 0.000121yea{r^{ - 1}}\end{align}\)

Rate constant of the reaction = 0.00012 year-1 \(\)

The time period taken it to 93.79% of the initial dose is as follows:

\(\begin{align}k &= {\bf{ }}\frac{{2.303}}{t}Log{\bf{ }}\frac{{\left( A \right)}}{{{{\left( A \right)}_0}}}\\t &= {\bf{ }}\frac{{2.303}}{k}Log{\bf{ }}\frac{{0.9379{{\left( A \right)}_0}}}{{{{\left( A \right)}_0}}}\\t &= {\bf{ }}\frac{{2.303}}{{0.000121}}Log{\bf{ }}64\\t &= {\bf{ }}\frac{{2.303}}{{0.000121}} \times 0.0115\\t &= {\bf{ }}218.88years\end{align}\)

Therefore, the time period was taken to kill Richard-III died in 218.88 years or 219 years.

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

Nitrogen monoxide reacts with chlorine according to the equation:

2NO(g) + Cl\({}_2\)(g)⟶ 2NOCl(g) The following initial rates of reaction have been observed for certain reactant concentrations:

What is the rate law that describes the rate’s dependence on the concentrations of NO and Cl2? What is the rate constant? What are the orders with respect to each reactant?

Chemical reactions occur when reactants collide. What are two factors that may prevent a collision from producing a chemical reaction?

The rate constant at 325°C for the decomposition reaction \({{\bf{C}}_{\bf{4}}}{{\bf{H}}_{\bf{8}}} \to {\bf{2}}{{\bf{C}}_{\bf{2}}}{{\bf{H}}_{\bf{4}}}\)is 6.1 × 10−8 s−1, and the activation energy is 261 kJ per mole of\({{\bf{C}}_{\bf{4}}}{{\bf{H}}_{\bf{8}}}\). Determine the frequency factor for the reaction.

Doubling the concentration of a reactant increases the rate of a reaction four times. With this knowledge, answer the following questions:

  1. What is the order of the reaction with respect to that reactant?
  2. Tripling the concentration of a different reactant increases the rate of a reaction three times. What is the order of the reaction with respect to that reactant?

The hydrolysis of the sugar sucrose to the sugars glucose and fructose, \({{\bf{C}}_{{\bf{12}}}}{{\bf{H}}_{{\bf{22}}}}{{\bf{O}}_{{\bf{11}}}}{\bf{ + }}{{\bf{H}}_{\bf{2}}}{\bf{O}} \to {{\bf{C}}_{\bf{6}}}{{\bf{H}}_{{\bf{12}}}}{{\bf{O}}_{\bf{6}}}{\bf{ + }}{{\bf{C}}_{\bf{6}}}{{\bf{H}}_{{\bf{12}}}}{{\bf{O}}_{\bf{6}}}\) follows a first-order rate equation for the disappearance of sucrose: \({\bf{Rate = k}}\left( {{{\bf{C}}_{{\bf{12}}}}{{\bf{H}}_{{\bf{22}}}}{{\bf{O}}_{{\bf{11}}}}} \right)\) (The products of the reaction, glucose and fructose, have the same molecular formulas but differ in the arrangement of the atoms in their molecules.)

  1. In neutral solution, \({\bf{k = 2}}{\bf{.1 \times 1}}{{\bf{0}}^{{\bf{ - 11}}}}{{\bf{s}}^{{\bf{ - 1}}}}\) at 27 °C and \({\bf{8}}{\bf{.5 \times 1}}{{\bf{0}}^{{\bf{ - 11}}}}{{\bf{s}}^{{\bf{ - 1}}}}\) at 37 °C. Determine the activation energy, the frequency factor, and the rate constant for this equation at 47 °C (assuming the kinetics remain consistent with the Arrhenius equation at this temperature).
  2. When a solution of sucrose with an initial concentration of 0.150 M reaches equilibrium, the concentration of sucrose is\({\bf{1}}{\bf{.65 \times 1}}{{\bf{0}}^{{\bf{ - 7}}}}{\bf{ M}}\). How long will it take the solution to reach equilibrium at 27 °C in the absence of a catalyst? Because the concentration of sucrose at equilibrium is so low, assume that the reaction is irreversible.
  3. Why does assuming that the reaction is irreversible simplify the calculation in part (b)?
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.