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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.

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Most popular questions from this chapter

The rate constant for the decomposition of acetaldehyde, \({\bf{C}}{{\bf{H}}_{\bf{3}}}{\bf{CHO}}\), to methane, \({\bf{C}}{{\bf{H}}_{\bf{4}}}\), and carbon monoxide, CO, in the gas phase is 1.1 ร— 10โˆ’2 L/mol/s at 703 K and 4.95 L/mol/s at 865 K. Determine the activation energy for this decomposition.

The rate law for the reaction: \({{\bf{H}}_{\bf{2}}}\) (g) + 2\({\bf{NO}}\) (g) โŸถ\({\bf{N}}2{\bf{O}}\) (g) + \({{\bf{H}}_{\bf{2}}}\)O(g) has been determined to be rate = k(\({\bf{NO}}\))2 (\({{\bf{H}}_{\bf{2}}}\)). What are the orders with respect to each reactant, and what is the overall order of the reaction?

How much and in what direction will each of the following effect the rate of the reaction:

CO(g) + \({\bf{NO}}{}_{\bf{2}}\) (g)โŸถ \({\bf{CO}}{}_{\bf{2}}\) (g) + NO(g) if the rate law for the reaction is rate =\({\bf{k(NO}}{}_{\bf{2}}{{\bf{)}}^{\bf{2}}}{\bf{a}}\)?

  1. Decreasing the pressure of \({\bf{NO}}{}_{\bf{2}}\) from 0.50 atm to 0.250 atm.
  2. Increasing the concentration of CO from 0.01 M to 0.03 M.

Experiments were conducted to study the rate of the reaction represented by this equation.(2)\({\rm{2NO(g) + 2}}{{\rm{H}}_{\rm{2}}}{\rm{(g) }} \to {\rm{ }}{{\rm{N}}_{\rm{2}}}{\rm{(g) + 2}}{{\rm{H}}_{\rm{2}}}{\rm{O(g)}}\)Initial concentrations and rates of reaction are given here.

Experiment Initial Concentration

\(\left( {{\bf{NO}}} \right){\rm{ }}\left( {{\bf{mol}}/{\bf{L}}} \right)\)

Initial Concentration, \(\left( {{{\bf{H}}_{\bf{2}}}} \right){\rm{ }}\left( {{\bf{mol}}/{\bf{L}}} \right)\)Initial Rate of Formation of \({{\bf{N}}_{\bf{2}}}{\rm{ }}\left( {{\bf{mol}}/{\bf{L}}{\rm{ }}{\bf{min}}} \right)\)
1\({\bf{0}}.{\bf{0060}}\)\({\bf{0}}.{\bf{00}}1{\bf{0}}\)\({\bf{1}}.{\bf{8}} \times {\bf{1}}{{\bf{0}}^{ - {\bf{4}}}}\)
2\({\bf{0}}.{\bf{0060}}\)\({\bf{0}}.{\bf{00}}2{\bf{0}}\)\({\bf{3}}.{\bf{6}} \times {\bf{1}}{{\bf{0}}^{ - {\bf{4}}}}\)
3\({\bf{0}}.{\bf{00}}1{\bf{0}}\)\({\bf{0}}.{\bf{0060}}\)\({\bf{0}}.{\bf{30}} \times {\bf{1}}{{\bf{0}}^{ - {\bf{4}}}}\)
4\({\bf{0}}.{\bf{00}}2{\bf{0}}\)\({\bf{0}}.{\bf{0060}}\)\({\bf{1}}.{\bf{2}} \times {\bf{1}}{{\bf{0}}^{ - {\bf{4}}}}\)

Consider the following questions:(a) Determine the order for each of the reactants, \({\bf{NO}}\) and \({{\bf{H}}_{\bf{2}}}\), from the data given and show your reasoning.(b) Write the overall rate law for the reaction.(c) Calculate the value of the rate constant, k, for the reaction. Include units.(d) For experiment 2, calculate the concentration of \({\bf{NO}}\)remaining when exactly one-half of the original amount of \({{\bf{H}}_{\bf{2}}}\) had been consumed.(e) The following sequence of elementary steps is a proposed mechanism for the reaction.Step 1:Step 2:Step 3:Based on the data presented, which of these is the rate determining step? Show that the mechanism is consistent with the observed rate law for the reaction and the overall stoichiometry of the reaction.

What is the half-life for the decomposition of \({{\bf{O}}_{\bf{3}}}\) when the concentration of \({{\bf{O}}_{\bf{3}}}\)is \({\bf{2}}{\bf{.35 \times 1}}{{\bf{0}}^{{\bf{ - 6}}}}\)M? The rate constant for this second-order reaction is 50.4 L/mol/h.

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