Reaction Orders, Integrated Rate Laws, and Half-Life

Reaction Orders

• Individual Reaction Order: Describes how the rate of a reaction depends on the concentration of a specific reactant.

• For a simple reaction $A \rightarrow \text{products}$, its order with respect to reactant A is determined by how its rate changes when $[A]$ changes:

  • First Order with respect to A: If the rate doubles when $[A]$ doubles, the rate depends on $[A]^1$.

  • Second Order with respect to A: If the rate quadruples when $[A]$ doubles, the rate depends on $[A]^2$.

  • Zero Order with respect to A: If the rate does not change when $[A]$ doubles, the rate does not depend on $[A]$ (i.e., depends on $[A]^0$).

Graphical Representation of Reaction Orders

• Plots of Reactant Concentration, $[A]$, versus Time: These graphs (visualized in Figure 16.8) show distinct curves for first-, second-, and zero-order reactions, illustrating the rate of decrease in reactant concentration over time.

• Plots of Rate versus Reactant Concentration, $[A]$: These graphs (visualized in Figure 16.9) depict the direct relationship between reaction rate and reactant concentration for different reaction orders.

Individual and Overall Reaction Orders

• For a general reaction, such as $2NO(g) + 2H<em>2(g) \rightarrow N</em>2(g) + 2H_2O(g)$, the rate law is experimentally determined.

  • Example: $Rate = k[NO]^2[H_2]^1$

  • Individual Orders:

    • Second order with respect to NO (exponent of $[NO]$ is 2).

    • First order with respect to $H<em>2$ (exponent of $[H</em>2]$ is 1).

  • Overall Order: The sum of the individual orders ($2 + 1 = 3$), making this reaction third order overall.

• Important Principle: Reaction orders must be determined from experimental data and cannot be deduced from the stoichiometric coefficients in the balanced chemical equation. For instance, $H_2$ has a coefficient of 2 in the balanced equation but a first-order dependence in the rate law.

Problem Solving: Determining Reaction Orders from Rate Laws (Sample Problem 16.2)

• Problem: Given a rate law, determine the reaction order with respect to each reactant and the overall order. Also, calculate the factor by which the rate changes if concentrations are altered.

• Plan: Inspect the exponents in the rate law to find individual orders, then sum them for the overall order. For rate changes, substitute the fold-change values into the rate law and calculate.

• Example (a): For $Rate = k[NO]^2[O_2]$

  • Second order with respect to NO.

  • First order with respect to $O_2$.

  • Third order overall.

  • If $[NO]$ triples and $[O<em>2]$ doubles: $Rate \text{ new} = k(3[NO])^2(2[O</em>2]) = k(9[NO]^2)(2[O<em>2]) = 18 \times k[NO]^2[O</em>2] = 18 \times Rate \text{ old}$. The rate increases by a factor of 18.

• Example (b): For $Rate = k[CH_3CHO]^2$

  • Second order in $CH_3CHO$.

  • Second order overall.

• Example (c): For $Rate = k[H<em>2O</em>2][I^-]$

  • First order in $H<em>2O</em>2$.

  • First order in $I^-$.

  • Zero order in $H^+$ (since $H^+$ does not appear in the rate law).

  • Second order overall.

Determining Reaction Orders from Experimental Data

• For a general reaction $A + 2B \rightarrow C + D$, the rate law is $Rate = k[A]^m[B]^n$.

• Method of Initial Rates: To determine $m$ and $n$, a series of experiments are run where:

  • One reactant's concentration is changed while others are kept constant.

  • The effect on the initial reaction rate is measured.

• Example: Reaction between $O<em>2$ and $NO$ ($2NO(g) + O</em>2(g) \rightarrow 2NO<em>2(g)$, with rate law $Rate = k[NO]^n[O</em>2]^m$).

  • Finding $m$ (order with respect to $O_2$):

    • Compare experiments where $[NO]$ is constant, but $[O_2]$ changes (e.g., experiments 1 and 2 from Table 16.1).

    • If $[O<em>2]$ doubles and the rate doubles, then $2^m = 2 \implies m = 1$. The reaction is first order with respect to $O</em>2$.

    • Mathematically: $rac{\text{Rate}<em>2}{\text{Rate}</em>1} = rac{k[NO]<em>2^n[O</em>2]<em>2^m}{k[NO]</em>1^n[O<em>2]</em>1^m} = rac{[O<em>2]</em>2^m}{[O<em>2]</em>1^m} = \bigg( rac{[O<em>2]</em>2}{[O<em>2]</em>1}\bigg)^m$

    • Solving for $m$: $m = rac{\log(\text{Rate}<em>2/\text{Rate}</em>1)}{\log([O<em>2]</em>2/[O<em>2]</em>1)}$. This is useful if the exponent is not easily found by inspection.

  • Finding $n$ (order with respect to $NO$):

    • Compare experiments where $[O_2]$ is constant, but $[NO]$ changes (e.g., experiments 1 and 3).

    • If $[NO]$ doubles and the rate quadruples, then $2^n = 4 \implies n = 2$. The reaction is second order with respect to NO.

• Types of Reaction Orders: Orders may be positive integers, zero, negative integers, or fractions.

• Once individual orders are determined, the full rate law can be written (e.g., $Rate = k[NO]^2[O_2]$).

Units of the Rate Constant (k)

• The value of $k$ is determined from experimental rate data.

• The units of $k$ depend on the overall reaction order.

Information Sequence to Determine Kinetic Parameters (Figure 16.10)

1. Initial Rates: Obtain initial rates from a series of plots of concentration versus time (by determining the slope of the tangent at $t=0$ for each plot).

2. Compare Initial Rates: Compare initial rates when one reactant concentration changes and others are held constant.

3. Reaction Orders: Determine individual reaction orders ($m, n$).

4. Rate Constant (k) and Actual Rate Law: Substitute initial rates, orders, and concentrations into the general rate law ($Rate = k[A]^m[B]^n$) and solve for $k$. This also reveals the actual rate law.

Problem Solving: Determining Reaction Orders & Rate Constants (Sample Problem 16.3)

• Problem: Use given initial rate data for $NO<em>2(g) + CO(g) \rightarrow NO(g) + CO</em>2(g)$ to find individual and overall reaction orders, calculate $k$, and then find the rate for new concentrations.

• Plan: (a) Use ratios of rate laws from experiments to find $m$ and $n$. (b) Substitute data from one experiment and calculated orders into the rate law to solve for $k$. (c) Use calculated $k$ and new concentrations to find the rate.

• Solution (a):

  • Order in $NO<em>2$ ($m$): Compare experiments 1 and 2 ($[CO]$ is constant). If $[NO</em>2]$ quadruples (from 0.10 M to 0.40 M) and the rate increases by a factor of 16 (0.0050 to 0.080 M/s), then $4^m = 16 \implies m = 2$. The reaction is second order in $NO_2$.

  • Order in CO ($n$): Compare experiments 1 and 3 ($[NO_2]$ is constant). If $[CO]$ doubles (0.10 M to 0.20 M) and the rate does not change (0.0050 M/s), then $2^n = 1 \implies n = 0$. The reaction is zero order in CO.

  • Rate Law: $Rate = k[NO<em>2]^2[CO]^0 = k[NO</em>2]^2$.

  • Overall Order: $2 + 0 = 2$ (second order overall).

• Solution (b) - Calculating k (using Experiment 1 data):

  • $0.0050 \text{ mol/L} \cdot \text{s} = k(0.10 \text{ mol/L})^2$

  • $k = rac{0.0050 \text{ mol/L} \cdot \text{s}}{(0.10 \text{ mol/L})^2} = rac{0.0050 \text{ mol/L} \cdot \text{s}}{0.0100 \text{ mol}^2/\text{L}^2} = 0.50 \text{ L/mol} \cdot \text{s}$.

• Solution (c) - Finding rate at new concentrations:

  • $Rate = (0.50 \text{ L/mol} \cdot \text{s})(0.65 \text{ mol/L})^2 = 0.21 \text{ mol/L} \cdot \text{s}$(when $[NO_2] = 0.65 \text{ M}$ and $[CO] = 0.25 \text{ M}$).

Problem Solving: Determining Reaction Orders from Molecular Scenes (Sample Problem 16.4)

• Problem: Given molecular scenes representing starting mixtures for experiments, determine reaction orders, write the rate law, and predict an initial rate.

• Plan: (a) Count particles (representing concentrations) and observe rate changes between experiments to determine orders. (b) Write the rate law using determined orders. (c) Use the rate law and particle counts for a new experiment to predict its rate.

• Solution (a): For reaction $A + B \rightarrow \text{products}$, with rate law $Rate = k[A]^x[B]^y$

  • Order with respect to A (red): Compare Expts 1 (2 A, 2 B, Rate = 0.060) and 2 (4 A, 2 B, Rate = 0.120). When $[A]$ doubles, Rate doubles ($2^x = 2$), so $x=1$. First order with respect to A.

  • Order with respect to B (blue): Compare Expts 1 (2 A, 2 B, Rate = 0.060) and 3 (2 A, 4 B, Rate = 0.240). When $[B]$ doubles, Rate quadruples ($2^y = 4$), so $y=2$. Second order with respect to B.

  • Overall Order: $1 + 2 = 3$ (third order overall).

• Solution (b) - Writing the rate law: $Rate = k[A][B]^2$

• Solution (c) - Predicting initial rate of Expt 4: Compare Expts 3 (2 A, 4 B, Rate = 0.240) and 4 (4 A, 4 B). $[A]$ doubles, while $[B]$ is constant. Since the reaction is first order in A, the rate doubles. $Rate<em>4 = 2 \times Rate</em>3 = 2 \times 0.240 = 0.480$

Integrated Rate Law for First-Order Reactions

• An integrated rate law includes time ($t$) as a variable, allowing calculation of concentrations at a specific time or the time required for a concentration change.

• Differential Rate Law: $Rate = - rac{d[A]}{dt} = k[A]$

• Integrated Rate Equation (Equation 16.4): $\$ln[A]<em>t = -kt + \ln[A]</em>0$ or $\$ln \frac{[A]<em>t}{[A]</em>0} = -kt$

  • $[A]_t$: concentration of A at time $t$

  • $[A]_0$: initial concentration of A (at $t=0$)

  • $k$: rate constant

Problem Solving: Determining Reactant Concentration After a Given Time in a First-Order Reaction (Sample Problem 16.5)

• Problem: For cyclobutane ($C<em>4H</em>8$) decomposition ($C<em>4H</em>8 \rightarrow 2C<em>2H</em>4$), a first-order reaction with $k = 87 \text{ s}^{-1}$.

  • (a) If $[C<em>4H</em>8]<em>0 = 2.00 \text{ M}$, what is $[C</em>4H<em>8]</em>t$ after $0.010 \text{ s}$?

  • (b) How long for $70.0\%$ of $C<em>4H</em>8$ to decompose?

• Plan: (a) Use the first-order integrated rate law to solve for $[A]_t$. (b) Determine the remaining concentration, then use the integrated rate law to solve for $t$.

• Solution (a):

  • $\ln[A]<em>t = -kt + \ln[A]</em>0$

  • $\ln[C<em>4H</em>8]_t = -(87 \text{ s}^{-1})(0.010 \text{ s}) + \ln(2.00 \text{ M})$

  • $\ln[C<em>4H</em>8]_t = -0.87 + 0.693 = -0.177$

  • $[C<em>4H</em>8]_t = e^{-0.177} = 0.838 \text{ M}$ (after taking the antilog).

• Solution (b):

  • If $70.0\%$ decomposes, $30.0\%$ remains. So, $[A]<em>t = 0.300 \times [A]</em>0 = 0.300 \times 2.00 \text{ M} = 0.600 \text{ M}$.

  • $\ln(0.600 \text{ M}) = -(87 \text{ s}^{-1})t + \ln(2.00 \text{ M})$

  • $-0.511 = -87t + 0.693$

  • $-1.204 = -87t \implies t = 0.0138 \text{ s}$.

Reaction Half-Life ($t_{1/2}$)

• Definition: The time taken for the concentration of a reactant to drop to half its initial value.

• For a First-Order Reaction (Equation 16.5):

  • Expression: $t_{1/2} = rac{\ln 2}{k} = rac{0.693}{k}$.

  • Key characteristic: For a first-order reaction, $t<em>{1/2}$ does not depend on the starting concentration ($[A]</em>0$). It is a constant at a particular temperature.

  • Relationship: Half-life and the rate constant ($k$) are inversely proportional.

  • Application: Radioactive decay is a first-order process, and its half-life is a useful indicator of nuclear stability.

• Visualizing Half-Lives (Figure 16.11): A plot of $[N<em>2O</em>5]$ versus time shows that the time interval to halve the concentration remains constant, regardless of the starting concentration for each interval.

Problem Solving: Using Molecular Scenes to Find Quantities at Various Times (Sample Problem 16.6)

• Problem: Substance A (green) decomposes to B (blue) and C (yellow) in a first-order gaseous reaction. Given molecular scenes at $t=0$ and $t=30.0 \text{ s}$, draw a scene at $t=60.0 \text{ s}$, find $k$, and calculate partial pressure of B at $t=90.0 \text{ s}$.

• Plan: (a) Determine half-life from given scenes, then predict particle counts at $t=60.0 \text{ s}$ (two half-lives). (b) Calculate $k$ from $t_{1/2}$. (c) Determine particle counts at $t=90.0 \text{ s}$ (three half-lives), calculate mole fraction of B, then its partial pressure.

• Solution (a) - Molecular Scene at 60.0 s:

  • At $t=0$: 8 A particles. At $t=30.0 \text{ s}$: 4 A particles, 4 B, 4 C. Thus, $t_{1/2} = 30.0 \text{ s}$.

  • At $t=60.0 \text{ s}$ (two half-lives): A will halve again. Number of A particles = $4/2 = 2$. Since each A $\rightarrow$ 1 B + 1 C, the decomposed A (6 particles) will form 6 B and 6 C particles. Added to the existing 4 B and 4 C, there will be 6 B and 6 C particles (from the first half-life) plus 2 B and 2 C (from the second half-life). So, the total number of B is $4+2=6$ and C is $4+2=6$. Ah, if 6 A decompose over two half lives, that means 4 A (from first half) + 2 A (from second half) = 6 A form 6 B + 6 C. So, at t=60s, there are 2 A, 6 B, 6 C particles. Correction from transcript solution: The solution says "particles of A form 6 of B and 6 of C" which results in 2 A, 6 B, 6 C for a total of 14 particles. Initial was 8 A. After one half life: 4 A, 4 B, 4 C (12 particles). After two half lives: 2 A. The 6 A that disappeared form 6 B and 6 C. So, it should be 2 A, 6 B, 6 C.

• Solution (b) - Finding k: $k = rac{\ln 2}{t_{1/2}} = rac{0.693}{30.0 \text{ s}} = 0.0231 \text{ s}^{-1}$.

• Solution (c) - Partial pressure of B at 90.0 s:

  • At $t=90.0 \text{ s}$ (three half-lives): Number of A particles = $8/2^3 = 1$. The 7 A particles that decomposed formed 7 B and 7 C. So, 1 A, 7 B, 7 C particles globally.

  • Total particles = $1 + 7 + 7 = 15$.

  • Mole fraction of B ($X_B$) = $rac{\text{mole B}}{\text{total moles}} = rac{7}{15}$.

  • Partial pressure of B ($P<em>B$) = $X</em>B \times P_{\text{total}} = rac{7}{15} \times 5.00 \text{ atm} = 2.33 \text{ atm}$.

Problem Solving: Determining the Half-Life of a First-Order Reaction (Sample Problem 16.7)

• Problem: Cyclopropane rearranges to propene (first-order reaction) with $k = 5.4 \times 10^{-2} \text{ s}^{-1}$ at $1000^{\circ}C$.

  • (a) What is $t_{1/2}$?

  • (b) How long for concentration to reach one-quarter of initial value?

• Plan: (a) Use the first-order half-life equation. (b) Recognize that one-quarter concentration means two half-lives.

• Solution (a):

  • $t_{1/2} = rac{0.693}{k} = rac{0.693}{5.4 \times 10^{-2} \text{ s}^{-1}} = 12.8 \text{ s}$.

• Solution (b):

  • Reaching one-quarter of the initial value means the concentration has been halved twice. This corresponds to $2 \times t_{1/2}$.

  • Time = $2 \times 12.8 \text{ s} = 25.6 \text{ s}$.

Integrated Rate Law and Half-Life for Second-Order Reactions

• For a Second-Order Reaction with one reactant (e.g., $A \rightarrow \text{products}$):

  • Differential Rate Law: $Rate = - rac{d[A]}{dt} = k[A]^2$

  • Integrated Rate Equation (Equation 16.6): $rac{1}{[A]<em>t} = kt + rac{1}{[A]</em>0}$

  • Half-Life (Equation 16.7): $t<em>{1/2} = rac{1}{k[A]</em>0}$

  • Key characteristic: For a second-order reaction, $t<em>{1/2}$ is inversely proportional to the initial reactant concentration ($[A]</em>0$). This means $t_{1/2}$ is not constant but changes as the reaction proceeds and $[A]$ decreases.

Problem Solving: Determining for Second-Order Reactions (Sample Problem 16.8)

• Problem: HI decomposes to $H<em>2$ and $I</em>2$ (second-order) with $k = 2.4 \times 10^{-2} \text{ L/mol} \cdot \text{s}$ at $25^{\circ}C$.

  • (a) If $0.0100 \text{ mol}$ HI in 1.0 L container, how long for $[HI]$ to reach $0.00900 \text{ mol/L}$?

  • (b) Calculate $t<em>{1/2}$ when $[HI]</em>0 = 1.50 \text{ mol/L}$ and when $[HI]_0 = 0.750 \text{ mol/L}$.

• Plan: (a) Use the second-order integrated rate law to solve for $t$. (b) Use the second-order half-life expression for each initial concentration.

• Solution (a):

  • $rac{1}{[HI]<em>t} = kt + rac{1}{[HI]</em>0}$

  • $rac{1}{0.00900 \text{ mol/L}} = (2.4 \times 10^{-2} \text{ L/mol} \cdot \text{s})t + rac{1}{0.0100 \text{ mol/L}}$

  • $111.1 \text{ L/mol} = (2.4 \times 10^{-2} \text{ L/mol} \cdot \text{s})t + 100.0 \text{ L/mol}$

  • $11.1 \text{ L/mol} = (2.4 \times 10^{-2} \text{ L/mol} \cdot \text{s})t$

  • $t = rac{11.1 \text{ L/mol}}{2.4 \times 10^{-2} \text{ L/mol} \cdot \text{s}} = 4.6 \times 10^{2} \text{ s}$.

• Solution (b):

  • When $[HI]<em>0 = 1.50 \text{ mol/L}$: $t</em>{1/2} = rac{1}{(2.4 \times 10^{-2} \text{ L/mol} \cdot \text{s})(1.50 \text{ mol/L})} = 28 \text{ s}$.

  • When $[HI]<em>0 = 0.750 \text{ mol/L}$: $t</em>{1/2} = rac{1}{(2.4 \times 10^{-2} \text{ L/mol} \cdot \text{s})(0.750 \text{ mol/L})} = 56 \text{ s}$.

  • This demonstrates that for a second-order reaction, reducing the initial concentration by half doubles the half-life.

Integrated Rate Law and Half-Life for Zero-Order Reactions

• For a Zero-Order Reaction (e.g., $A \rightarrow \text{products}$):

  • Differential Rate Law: $Rate = - rac{d[A]}{dt} = k$

  • Integrated Rate Equation (Equation 16.8): $[A]<em>t = -kt + [A]</em>0$

  • Half-Life: $t<em>{1/2} = rac{[A]</em>0}{2k}$

  • Key characteristic: For a zero-order reaction, $t<em>{1/2}$ is directly proportional to the initial reactant concentration ($[A]</em>0$). As $[A]<em>0$ decreases, $t</em>{1/2}$ also decreases.

Graphical Method for Finding Reaction Order from Integrated Rate Laws

• The integrated rate laws can be rearranged into the form of a straight line ($y = mx + b$) to graphically determine the reaction order.

• First-Order Reactions (Figure 16.12A):

  • Integrated rate law: $\$ln[A]<em>t = -kt + \ln[A]</em>0$

  • Straight-line form: Plot $\$ln[A]_t$ (y-axis) versus time ($t$, x-axis).

  • Slope: $-k$

  • Y-intercept: $\$ln[A]_0$

  • A linear plot indicates a first-order reaction.

• Second-Order Reactions (Figure 16.12B):

  • Integrated rate law: $rac{1}{[A]<em>t} = kt + rac{1}{[A]</em>0}$

  • Straight-line form: Plot $rac{1}{[A]_t}$ (y-axis) versus time ($t$, x-axis).

  • Slope: $k$

  • Y-intercept: $rac{1}{[A]_0}$

  • A linear plot indicates a second-order reaction.

• Zero-Order Reactions (Figure 16.12C):

  • Integrated rate law: $[A]<em>t = -kt + [A]</em>0$

  • Straight-line form: Plot $[A]_t$ (y-axis) versus time ($t$, x-axis).

  • Slope: $-k$

  • Y-intercept: $[A]_0$

  • A linear plot indicates a zero-order reaction.

• Graphical Determination Example: Decomposition of $N<em>2O</em>5$ (Figure 16.13 and Table 16.3):

  • Concentration data for $[N<em>2O</em>5]$ at various times is transformed into $\$ln[N<em>2O</em>5]$ and $rac{1}{[N<em>2O</em>5]}$.

  • By plotting $[N<em>2O</em>5]$ vs. $t$, $\$ln[N<em>2O</em>5]$ vs. $t$, and $rac{1}{[N<em>2O</em>5]}$ vs. $t$, the plot that yields a straight line reveals the reaction order.

  • For $N<em>2O</em>5$ decomposition, the plot of $\$ln[N<em>2O</em>5]$ versus time gives a straight line, confirming it is a first-order reaction.

Overview of Zero-Order, First-Order, and Simple Second-Order Reactions (Table 16.3)