6.4 Comprehensive Guide to Integrated Rate Laws and Half-Life Equations

Introduction to Integrated Rate Laws

• Definition: Integrated rate laws are mathematical expressions that relate the concentration of reactants to the amount of time ($t$) that has transpired since the beginning of a chemical reaction.
• Comparison to Differential Rate Laws:
  • Differential Rate Laws: Relate the rate of reaction to the concentration of reactants (e.g., $\text{Rate} = k[A]^n$).
  • Integrated Rate Laws: Relate the specific concentration of a reactant at a certain time to the initial concentration and elapsed time.
• Utility: These equations are essential for:
  • Determining the amount of reactant or product present after a specific duration.
  • Estimating the time required to produce a specific amount of product.
  • Identifying the order of a reaction using concentration/time data.

First-Order Reactions

• Differential Rate Law: $\text{Rate} = k[A]^1$
• Integrated Rate Law Formulations:
  • Exponential form: $[A]_t = [A]_0 e^{-kt}$
  • Logarithmic form: $\ln([A]_t) = \ln([A]_0) - kt$
  • Rearranged ratio form: $\ln\left(\frac{[A]_0}{[A]_t}\right) = kt$
• Variables:
  • $[A]_t$: Concentration of reactant $A$ at time $t$.
  • $[A]_0$: Initial concentration of reactant $A$.
  • $k$: Rate constant.
  • $t$: Time elapsed.

Guided Practice: First-Order Decay of Iodine-131

• Problem: Iodine-131 undergoes radioactive decay to form Xenon-131. The reaction is first-order with a rate constant ($k$) of $0.138 \, \text{day}^{-1}$. Calculate the time required for $90\%$ of the Iodine in a $0.5 \, \text{M}$ solution to decay.
• Analysis:
  • Initial concentration ($[A]_0$): $0.5 \, \text{M}$.
  • Percent decayed: $90\%$.
  • Percent remaining: $10\%$.
  • Final concentration ($[A]_t$): $0.1 \times 0.5 = 0.05 \, \text{M}$.
• Calculation:
  • Equation: $t = \frac{\ln\left(\frac{[A]_0}{[A]_t}\right)}{k}$
  • Substituting values: $t = \frac{\ln\left(\frac{0.5}{0.05}\right)}{0.138 \, \text{day}^{-1}} = \frac{\ln(10)}{0.138 \, \text{day}^{-1}}$
  • Raw result: $16.68539922 \, \text{days}$.
  • Final Answer: Rounded to three significant figures, it takes $16.7 \, \text{days}$.

Second-Order Reactions

• Differential Rate Law: $\text{Rate} = k[A]^2$
• Integrated Rate Law: $\frac{1}{[A]_t} = kt + \frac{1}{[A]_0}$

Guided Practice: Dimerization of 1,3-Butadiene

• Problem: The initial concentration of 1,3-butadiene is $0.0200 \, \text{M}$. Calculate the concentration remaining after $20.0 \, \text{minutes}$ if the rate constant ($k$) is $5.76 \times 10^{-2} \, \text{L} \, \text{mol}^{-1} \, \text{min}^{-1}$.
• Step 1: Simplify the right hand side:
  • $kt + \frac{1}{[A]_0} = (5.76 \times 10^{-2} \, \text{L} \, \text{mol}^{-1} \, \text{min}^{-1} \times 20.0 \, \text{min}) + \frac{1}{0.0200 \, \text{M}}$
  • $kt = 1.152 \, \text{L} \, \text{mol}^{-1}$
  • $\frac{1}{[A]_0} = 50 \, \text{M}^{-1}$
  • Total sum: $1.152 + 50 = 51.152 \, \text{M}^{-1}$
• Step 2: Solve for $[A]_t$:
  • $\frac{1}{[A]_t} = 51.152$
  • $[A]_t = \frac{1}{51.152} = 0.019549577 \, \text{M}$.
• Final Answer: Rounded to three significant figures, the concentration is $0.0195 \, \text{mol} \, \text{L}^{-1}$.

Zero-Order Reactions

• Differential Rate Law: $\text{Rate} = k$
• Note: The rate is constant and independent of the reactant concentration, assuming certain conditions are met.
• Integrated Rate Law: $[A]_t = -kt + [A]_0$

Guided Practice: Zero-Order Decomposition of Ammonia

• Problem: A zero-order plot shows an initial ammonia concentration of $0.0028 \, \text{mol} \, \text{L}^{-1}$. At what time (in minutes) will the concentration reach $0.0001 \, \text{mol} \, \text{L}^{-1}$? (Rate constant $k = 1.3 \times 10^{-6} \, \text{mol} \, \text{L}^{-1} \, \text{s}^{-1}$).
• Calculation:
  • Rearranged for $t$: $t = \frac{[A]_t - [A]_0}{-k}$
  • Substituting: $t = \frac{0.0001 - 0.0028}{-1.3 \times 10^{-6}}$
  • Result in seconds: $2076.923077 \, \text{s}$.
  • Conversion to minutes: $\frac{2076.923077}{60} = 34.61538462 \, \text{min}$.
• Final Answer: Rounded to two significant figures, the time is $35 \, \text{minutes}$.

Concepts of Half-Life ($t_{1/2}$

• Definition: The time required for one half of a given amount of reactant to be consumed.
• Sequential Half-Lives: In each succeeding half-life, half of the remaining concentration is consumed.
  • Example: Starting with $1.000 \, \text{M}$. After one half-life ($6 \, \text{hours}$), concentration is $0.500 \, \text{M}$. After a second half-life ($12 \, \text{hours}$ total), concentration is $0.250 \, \text{M}$.

Half-Life Equations by Reaction Order

• First-Order: $t_{1/2} = \frac{0.693}{k}$
  • The half-life is independent of initial concentration.
  • Inverse relationship: As $k$ increases (faster reaction), $t_{1/2}$ decreases.
• Second-Order: $t_{1/2} = \frac{1}{k[A]_0}$
  • The half-life is inversely proportional to the initial concentration.
  • $t_{1/2}$ changes as the reaction proceeds because it depends on the concentration at the start of that interval.
• Zero-Order: $t_{1/2} = \frac{[A]_0}{2k}$
  • The half-life is directly proportional to the initial concentration and inversely proportional to the rate constant.

Practice Problem: Half-Life Calculation for Ammonia

• Reaction Type: Zero-order thermal decomposition of ammonia on tungsten.
• Data: $k = 1.3 \times 10^{-6} \, \text{mol} \, \text{L}^{-1} \, \text{s}^{-1}$; $[A]_0 = 0.0028 \, \text{mol} \, \text{L}^{-1}$.
• Calculation:
  • $t_{1/2} = \frac{0.0028}{2 \times (1.3 \times 10^{-6}) } = 1076.923077 \, \text{s}$.
  • Conversion to minutes: $\frac{1076.923077}{60} = 17.9487 \, \text{min}$.
• Final Answer: Rounded to two significant figures, the half-life is $18 \, \text{minutes}$.

Questions & Discussion

• Dialogue Interlude:
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