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A sample of Uranium-238 has a half-life of 4.5 imes 109 years. (a) What is the half-life of Uranium-238? (b) Write the nuclear equation for its alpha decay. (c) If you start with 5 g of Uranium-238, h

Physics

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Here are the solutions to the questions:

a) i) The half-life of 4.5×1094.5 \times 10^9 years means that it takes 4.5×1094.5 \times 10^9 years for half of the initial amount of a radioactive sample of Uranium-238 to decay.

a) ii) The radioactive decay equation for Uranium-238 by alpha emission is:

92238U→90234Th+24He^{238}_{92}\text{U} \to ^{234}_{90}\text{Th} + ^4_2\text{He}

a) iii)

  1. Identify the given values and the decay formula.

    Initial mass (N0N_0) = 5 g5 \text{ g}
    Final mass (NN) = 10%10\% of 5 g=0.10×5 g=0.5 g5 \text{ g} = 0.10 \times 5 \text{ g} = 0.5 \text{ g}
    Half-life (t1/2t_{1/2}) = 4.5×109 years4.5 \times 10^9 \text{ years}
    The radioactive decay formula is N=N0(12)t/t1/2N = N_0 \left(\frac{1}{2}\right)^{t/t_{1/2}}.

  2. Substitute the values into the formula.

    0.5=5(12)t/(4.5×109)0.5 = 5 \left(\frac{1}{2}\right)^{t/(4.5 \times 10^9)}
  3. Isolate the exponential term.

    0.55=(12)t/(4.5×109)\frac{0.5}{5} = \left(\frac{1}{2}\right)^{t/(4.5 \times 10^9)} 0.1=(12)t/(4.5×109)0.1 = \left(\frac{1}{2}\right)^{t/(4.5 \times 10^9)}
  4. Take the natural logarithm of both sides.

    ln⁡(0.1)=ln⁡((12)t/(4.5×109))\ln(0.1) = \ln\left(\left(\frac{1}{2}\right)^{t/(4.5 \times 10^9)}\right) ln⁡(0.1)=t4.5×109ln⁡(0.5)\ln(0.1) = \frac{t}{4.5 \times 10^9} \ln(0.5)
  5. Solve for tt.

    t=ln⁡(0.1)ln⁡(0.5)×4.5×109 yearst = \frac{\ln(0.1)}{\ln(0.5)} \times 4.5 \times 10^9 \text{ years} t=−2.302585−0.693147×4.5×109 yearst = \frac{-2.302585}{-0.693147} \times 4.5 \times 10^9 \text{ years} t≈3.3219×4.5×109 yearst \approx 3.3219 \times 4.5 \times 10^9 \text{ years} t≈1.495×1010 yearst \approx \textbf{1.495} \times \textbf{10}^{\textbf{10}} \textbf{ years}

    a) iv) A. The same element can have different atomic masses because of the existence of isotopes. Isotopes are atoms of the same element that have the same number of protons (and thus the same atomic number) but different numbers of neutrons, leading to different mass numbers.

    a) iv) B.

  6. Use the formula for relative atomic mass (RAM).

    RAM=∑(mass of isotope×relative abundance)100\text{RAM} = \sum \frac{(\text{mass of isotope} \times \text{relative abundance})}{100}
  7. Substitute the given data for Barium.

    RAM=(134×2.2)+(135×5.5)+(136×7.7)+(137×12.1)+(138×72.5)100\text{RAM} = \frac{(134 \times 2.2) + (135 \times 5.5) + (136 \times 7.7) + (137 \times 12.1) + (138 \times 72.5)}{100}
  8. Calculate the sum of the products.

    RAM=294.8+742.5+1047.2+1657.7+10005100\text{RAM} = \frac{294.8 + 742.5 + 1047.2 + 1657.7 + 10005}{100} RAM=13747.2100\text{RAM} = \frac{13747.2}{100}
  9. Calculate the relative atomic mass.

    RAM=137.472\text{RAM} = \textbf{137.472}

    b) i) A rate law is an equation that expresses the relationship between the rate of a chemical reaction and the concentrations of the reactants, raised to appropriate powers (reaction orders), at a specific temperature.

    b) ii)

  10. Determine the order with respect to each reactant based on the given information.

    • Doubling the concentration of A2\text{A}_2 doubles the reaction rate, so the reaction is first order with respect to A2\text{A}_2. (Rate∝[A2]1\text{Rate} \propto [\text{A}_2]^1)
    • Tripling the concentration of B\text{B} triples the reaction rate, so the reaction is first order with respect to B\text{B}. (Rate∝[B]1\text{Rate} \propto [\text{B}]^1)
    • Doubling the concentration of C\text{C} has no effect on the rate, so the reaction is zero order with respect to C\text{C}. (Rate∝[C]0\text{Rate} \propto [\text{C}]^0)
  11. Combine the orders to write the rate law.

    Rate=k[A2]1[B]1[C]0\text{Rate} = k[\text{A}_2]^1[\text{B}]^1[\text{C}]^0 Rate=k[A2][B]\text{Rate} = \boxed{k[\text{A}_2][\text{B}]}

    b) iii) A mechanism consistent with the rate law Rate=k[A2][B]\text{Rate} = k[\text{A}_2][\text{B}] would have the slowest (rate-determining) step involving one molecule of A2\text{A}_2 and one molecule of B\text{B}. Reactant C\text{C} is not involved in the rate-determining step.

    Proposed Mechanism:
    Step 1 (Slow): A2+B→AB+A\text{A}_2 + \text{B} \to \text{AB} + \text{A} (Rate-determining step)
    Step 2 (Fast): A+C→AC\text{A} + \text{C} \to \text{AC}
    Overall Reaction: A2+B+C→AB+AC\text{A}_2 + \text{B} + \text{C} \to \text{AB} + \text{AC}

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