Solution:

Given reaction:

$4NH3\text{(g)} + 5O2\text{(g)} \rightarrow 4NO\text{(g)} + 6H2O\text{(g)}$

Given: O₂ consumed at 2.0 × 10⁻³ M/s

Note: "Consumed" means disappearance, so Δ[O₂]/Δt is negative

(a) Rate of reaction

General rate expression:

For reaction: aA + bB → cC + dD

$\text{Rate} = -\frac{1}{a}\frac{\Delta[A]}{\Delta t} = -\frac{1}{b}\frac{\Delta[B]}{\Delta t} = +\frac{1}{c}\frac{\Delta[C]}{\Delta t} = +\frac{1}{d}\frac{\Delta[D]}{\Delta t}$

For our reaction:

$\text{Rate} = -\frac{1}{4}\frac{\Delta[NH_3]}{\Delta t} = -\frac{1}{5}\frac{\Delta[O_2]}{\Delta t} = +\frac{1}{4}\frac{\Delta[NO]}{\Delta t} = +\frac{1}{6}\frac{\Delta[H_2O]}{\Delta t}$

Using O₂ term:

Given: O₂ consumed at 2.0 × 10⁻³ M/s

$\frac{\Delta[O_2]}{\Delta t} = -2.0 \times 10^{-3} \text{ M/s}$

(Negative because O₂ is disappearing)

Calculate rate:

$\text{Rate} = -\frac{1}{5}\frac{\Delta[O_2]}{\Delta t}$

$\text{Rate} = -\frac{1}{5}(-2.0 \times 10^{-3})$

$\text{Rate} = \frac{2.0 \times 10^{-3}}{5}$

$\text{Rate} = 0.40 \times 10^{-3} = 4.0 \times 10^{-4} \text{ M/s}$

Answer (a):

$\boxed{\text{Rate} = 4.0 \times 10^{-4} \text{ M/s}}$

(b) Rate of NO production

Using rate expression:

$\text{Rate} = +\frac{1}{4}\frac{\Delta[NO]}{\Delta t}$

Solve for Δ[NO]/Δt:

$\frac{\Delta[NO]}{\Delta t} = 4 \times \text{Rate}$

$\frac{\Delta[NO]}{\Delta t} = 4 \times (4.0 \times 10^{-4})$

$\frac{\Delta[NO]}{\Delta t} = 16 \times 10^{-4}$

$\frac{\Delta[NO]}{\Delta t} = 1.6 \times 10^{-3} \text{ M/s}$

Answer (b):

$\boxed{\frac{\Delta[NO]}{\Delta t} = 1.6 \times 10^{-3} \text{ M/s}}$

Interpretation: NO produced at 1.6 × 10⁻³ M/s (positive = production)

(c) Rate of NH₃ consumption

Using rate expression:

$\text{Rate} = -\frac{1}{4}\frac{\Delta[NH_3]}{\Delta t}$

Solve for Δ[NH₃]/Δt:

$\frac{\Delta[NH_3]}{\Delta t} = -4 \times \text{Rate}$

$\frac{\Delta[NH_3]}{\Delta t} = -4 \times (4.0 \times 10^{-4})$

$\frac{\Delta[NH_3]}{\Delta t} = -16 \times 10^{-4}$

$\frac{\Delta[NH_3]}{\Delta t} = -1.6 \times 10^{-3} \text{ M/s}$

Answer (c):

$\boxed{\frac{\Delta[NH_3]}{\Delta t} = -1.6 \times 10^{-3} \text{ M/s}}$

Or, stating as consumption rate (magnitude):

$\boxed{\text{NH}_3 \text{ consumed at } 1.6 \times 10^{-3} \text{ M/s}}$

Summary of Results

Verification

Check stoichiometric relationships:

From balanced equation: 4 NH₃ : 5 O₂ : 4 NO : 6 H₂O

From our rates:

$\frac{|\Delta[NH_3]/\Delta t|}{4} = \frac{1.6 \times 10^{-3}}{4} = 4.0 \times 10^{-4}$ ✓

$\frac{|\Delta[O_2]/\Delta t|}{5} = \frac{2.0 \times 10^{-3}}{5} = 4.0 \times 10^{-4}$ ✓

$\frac{|\Delta[NO]/\Delta t|}{4} = \frac{1.6 \times 10^{-3}}{4} = 4.0 \times 10^{-4}$ ✓

All equal to the rate! ✓

Key Insights

1. Stoichiometric relationships:

• NH₃ and NO have same coefficient (4)
• Therefore consumed/produced at same rate
• Both: 1.6 × 10⁻³ M/s

2. Ratio comparison:

$\frac{\Delta[NH_3]/\Delta t}{\Delta[O_2]/\Delta t} = \frac{-1.6 \times 10^{-3}}{-2.0 \times 10^{-3}} = \frac{1.6}{2.0} = \frac{4}{5}$

Matches stoichiometry: 4 NH₃ : 5 O₂ ✓

3. Why divide by coefficients?

• Makes "rate of reaction" unique
• Same value regardless of which species you measure
• Accounts for stoichiometric ratios

4. Sign conventions:

• Reactants: negative Δ[]/Δt (decreasing)
• Products: positive Δ[]/Δt (increasing)
• Rate of reaction: always positive

Alternative approach:

Could also use ratios directly:

Given: Δ[O₂]/Δt = -2.0 × 10⁻³ M/s

Want: Δ[NH₃]/Δt

From stoichiometry: 4 NH₃ consumed per 5 O₂

$\frac{\Delta[NH_3]/\Delta t}{\Delta[O_2]/\Delta t} = \frac{4}{5}$

$\frac{\Delta[NH_3]}{\Delta t} = \frac{4}{5} \times \frac{\Delta[O_2]}{\Delta t}$

$\frac{\Delta[NH_3]}{\Delta t} = \frac{4}{5} \times (-2.0 \times 10^{-3})$

$\frac{\Delta[NH_3]}{\Delta t} = -1.6 \times 10^{-3} \text{ M/s}$ ✓

Same answer!

This is the Ostwald process:

Industrial production of nitric acid:

Step 1: 4NH₃ + 5O₂ → 4NO + 6H₂O (this reaction)

Step 2: 2NO + O₂ → 2NO₂

Step 3: 3NO₂ + H₂O → 2HNO₃ + NO

Used to make:

• Fertilizers
• Explosives
• Various chemicals

Conditions:

• High temperature (850-900°C)
• Platinum-rhodium catalyst
• Important industrial process

Solution:

(a) Rate law: Rate = k[A]^m[B]^n

Find m (order with respect to A): Compare Experiments 1 and 2 ([B] constant):

• [A] doubles: 0.10 → 0.20
• Rate doubles: 0.015 → 0.030
• Therefore: 2^m = 2, so m = 1

Find n (order with respect to B): Compare Experiments 1 and 3 ([A] constant):

• [B] doubles: 0.10 → 0.20
• Rate quadruples: 0.015 → 0.060
• Therefore: 2^n = 4, so n = 2

Rate law: Rate = k[A][B]²

(b) Rate constant: Using Experiment 1: 0.015 = k(0.10)(0.10)² 0.015 = k(0.0010) k = 15 M⁻²s⁻¹

(c) Overall order: m + n = 1 + 2 = 3rd order (first order in A, second order in B)

Solution:

Given reaction:

$2NO\text{(g)} + Cl2\text{(g)} \rightarrow 2NOCl\text{(g)}$

Data table:

General rate law form:

$\text{Rate} = k[NO]^m[Cl_2]^n$

Need to find: m, n, k

(a) Determine rate law

Find order with respect to Cl₂ (n):

Compare Exp 1 and 2 (keep [NO] constant, vary [Cl₂])

Set up ratio:

$\frac{\text{rate}_2}{\text{rate}_1} = \frac{k[NO]_2^m[Cl_2]_2^n}{k[NO]_1^m[Cl_2]_1^n}$

Since [NO] is same in both:

$\frac{\text{rate}_2}{\text{rate}_1} = \frac{[Cl_2]_2^n}{[Cl_2]_1^n} = \left(\frac{[Cl_2]_2}{[Cl_2]_1}\right)^n$

Substitute values:

$\frac{0.36}{0.18} = \left(\frac{0.20}{0.10}\right)^n$

$2 = 2^n$

$n = 1$

First order in Cl₂

Find order with respect to NO (m):

Compare Exp 2 and 3 (keep [Cl₂] constant, vary [NO])

Set up ratio:

$\frac{\text{rate}_3}{\text{rate}_2} = \frac{k[NO]_3^m[Cl_2]_3^n}{k[NO]_2^m[Cl_2]_2^n}$

Since [Cl₂] is same in both:

$\frac{\text{rate}_3}{\text{rate}_2} = \frac{[NO]_3^m}{[NO]_2^m} = \left(\frac{[NO]_3}{[NO]_2}\right)^m$

Substitute values:

$\frac{1.44}{0.36} = \left(\frac{0.20}{0.10}\right)^m$

$4 = 2^m$

$2^2 = 2^m$

$m = 2$

Second order in NO

Rate law:

$\boxed{\text{Rate} = k[NO]^2[Cl_2]}$

Orders:

• With respect to NO: 2 (second order)
• With respect to Cl₂: 1 (first order)
• Overall: 2 + 1 = 3 (third order)

(b) Calculate k with units

Use any experiment; let's use Exp 1:

$\text{Rate} = k[NO]^2[Cl_2]$

Solve for k:

$k = \frac{\text{Rate}}{[NO]^2[Cl_2]}$

Substitute values from Exp 1:

$k = \frac{0.18}{(0.10)^2(0.10)}$

$k = \frac{0.18}{(0.01)(0.10)}$

$k = \frac{0.18}{0.001}$

$k = 180$

Determine units of k:

From rate equation:

$\text{Rate (M/s)} = k[NO]^2[Cl_2]$

$\text{M/s} = k \times \text{M}^2 \times \text{M}$

$\text{M/s} = k \times \text{M}^3$

Solve for units of k:

$k = \frac{\text{M/s}}{\text{M}^3} = \frac{\text{M}}{\text{s} \cdot \text{M}^3} = \frac{1}{\text{M}^2 \cdot \text{s}} = \text{M}^{-2}\text{s}^{-1}$

Answer (b):

$\boxed{k = 180 \text{ M}^{-2}\text{s}^{-1}}$

Or equivalently: k = 180 L²·mol⁻²·s⁻¹

Verify with other experiments:

Check with Exp 2:

$k = \frac{0.36}{(0.10)^2(0.20)} = \frac{0.36}{0.002} = 180$ ✓

Check with Exp 3:

$k = \frac{1.44}{(0.20)^2(0.20)} = \frac{1.44}{0.008} = 180$ ✓

Consistent! ✓

(c) Calculate rate at [NO] = 0.050 M, [Cl₂] = 0.020 M

Use rate law with k = 180 M⁻²s⁻¹:

$\text{Rate} = k[NO]^2[Cl_2]$

$\text{Rate} = 180 \times (0.050)^2 \times (0.020)$

Calculate (0.050)²:

$(0.050)^2 = 0.0025$

Calculate rate:

$\text{Rate} = 180 \times 0.0025 \times 0.020$

$\text{Rate} = 180 \times 0.000050$

$\text{Rate} = 0.0090 \text{ M/s}$

$\text{Rate} = 9.0 \times 10^{-3} \text{ M/s}$

Answer (c):

$\boxed{\text{Rate} = 9.0 \times 10^{-3} \text{ M/s} \text{ or } 0.0090 \text{ M/s}}$

Summary of Results

Key Observations

1. Rate law vs stoichiometry:

Balanced equation: 2NO + Cl₂ → 2NOCl

Rate law: Rate = k[NO]²[Cl₂]

Notice:

• Exponent for NO: 2 (matches coefficient)
• Exponent for Cl₂: 1 (matches coefficient)

This is coincidental!

• Rate law determined experimentally
• Sometimes matches, sometimes doesn't
• Never assume exponents = coefficients

2. Reaction order interpretation:

Second order in NO:

• If [NO] doubles, rate quadruples
• NO concentration has large effect
• Suggests 2 NO molecules in rate-determining step

First order in Cl₂:

• If [Cl₂] doubles, rate doubles
• Linear dependence
• Suggests 1 Cl₂ molecule in rate-determining step

Third order overall:

• Complex reaction
• Likely multi-step mechanism
• Rate-determining step involves 2 NO + 1 Cl₂

3. Initial rates method steps:

Step 1: Identify pairs where only one concentration varies

Step 2: Set up rate ratio equation

Step 3: Concentrations constant cancel out

Step 4: Solve for exponent

Step 5: Repeat for each reactant

Step 6: Calculate k from any experiment

Step 7: Verify k consistent across all experiments

Units of k for different orders:

Our reaction: Order = 3, so k has units M⁻²·s⁻¹ ✓

Real-world context:

This reaction:

• Gas-phase radical reaction
• Studied extensively for mechanism
• Shows typical behavior for NO reactions

Proposed mechanism:

Step 1 (fast equilibrium):

$NO + Cl2 \rightleftharpoons NOCl2$

Step 2 (slow, rate-determining):

$NOCl2 + NO \rightarrow 2NOCl$

Overall: 2NO + Cl₂ → 2NOCl ✓

Rate-determining step involves:

• 1 NOCl₂ (which comes from 1 NO + 1 Cl₂)
• 1 NO
• Total: 2 NO + 1 Cl₂

This explains the rate law: Rate = k[NO]²[Cl₂] ✓

Practice tip:

When doing initial rates problems:

1. ✓ Organize data in table
2. ✓ Find pairs where only ONE concentration changes
3. ✓ Set up ratio (things that don't change cancel)
4. ✓ Solve for exponent
5. ✓ Calculate k from any experiment
6. ✓ Check k is same for all experiments
7. ✓ Pay attention to units!

Solution:

(a) Rate law: Rate = k[A]¹[B]² or Rate = k[A][B]²

(b) Rate change factor: Initial rate: Rate₁ = k[A]₁[B]₁² New rate: Rate₂ = k[A]₂[B]₂²

Where: [A]₂ = 3[A]₁ and [B]₂ = 2[B]₁

Rate₂ = k(3[A]₁)(2[B]₁)² Rate₂ = k(3[A]₁)(4[B]₁²) Rate₂ = 12k[A]₁[B]₁² Rate₂ = 12 × Rate₁

The rate increases by a factor of 12

(c) Units of k: From Rate = k[A][B]² M/s = k × M × M² M/s = k × M³ k = M/s ÷ M³ = M⁻²s⁻¹

General pattern for units:

• 0th order: M/s
• 1st order: s⁻¹
• 2nd order: M⁻¹s⁻¹
• 3rd order: M⁻²s⁻¹

Solution:

Given:

• Reaction: 2H₂O₂(aq) → 2H₂O(l) + O₂(g)
• Order: First order
• k = 1.8 × 10⁻⁵ s⁻¹ at 20°C
• [H₂O₂]₀ = 0.30 M

First order integrated rate law:

$\ln[A]_t = \ln[A]_0 - kt$

Or:

$[A]_t = [A]_0 e^{-kt}$

(a) Concentration after 1.0 hour

Given:

• t = 1.0 hour
• [H₂O₂]₀ = 0.30 M

Convert time to seconds:

$t = 1.0 \text{ hour} \times \frac{60 \text{ min}}{1 \text{ hour}} \times \frac{60 \text{ s}}{1 \text{ min}} = 3600 \text{ s}$

Use integrated rate law:

$\ln[H_2O_2]_t = \ln[H_2O_2]_0 - kt$

Substitute values:

$\ln[H_2O_2]_t = \ln(0.30) - (1.8 \times 10^{-5})(3600)$

Calculate each term:

$\ln(0.30) = -1.204$

$(1.8 \times 10^{-5})(3600) = 0.0648$

Substitute:

$\ln[H_2O_2]_t = -1.204 - 0.0648$

$\ln[H_2O_2]_t = -1.269$

Take exponential:

$[H_2O_2]_t = e^{-1.269}$

$[H_2O_2]_t = 0.281 \text{ M}$

Answer (a):

$\boxed{[H_2O_2]_t = 0.28 \text{ M}}$

Alternative method using exponential form:

$[H_2O_2]_t = [H_2O_2]_0 e^{-kt}$

$[H_2O_2]_t = 0.30 \times e^{-(1.8 \times 10^{-5})(3600)}$

$[H_2O_2]_t = 0.30 \times e^{-0.0648}$

$e^{-0.0648} = 0.937$

$[H_2O_2]_t = 0.30 \times 0.937 = 0.281 \text{ M}$

Same answer! ✓

Interpretation:

• Initial: 0.30 M
• After 1 hour: 0.28 M
• Decrease: 0.02 M (about 6.3% decomposed)
• Very slow reaction at 20°C without catalyst

(b) Time to drop from 0.30 M to 0.10 M

Given:

• [H₂O₂]₀ = 0.30 M
• [H₂O₂]_t = 0.10 M
• Find: t

Use integrated rate law:

$\ln[H_2O_2]_t = \ln[H_2O_2]_0 - kt$

Rearrange to solve for t:

$kt = \ln[H_2O_2]_0 - \ln[H_2O_2]_t$

$kt = \ln\left(\frac{[H_2O_2]_0}{[H_2O_2]_t}\right)$

$t = \frac{1}{k}\ln\left(\frac{[H_2O_2]_0}{[H_2O_2]_t}\right)$

Substitute values:

$t = \frac{1}{1.8 \times 10^{-5}}\ln\left(\frac{0.30}{0.10}\right)$

$t = \frac{1}{1.8 \times 10^{-5}}\ln(3.0)$

Calculate ln(3.0):

$\ln(3.0) = 1.099$

Calculate t:

$t = \frac{1.099}{1.8 \times 10^{-5}}$

$t = 6.106 \times 10^4 \text{ s}$

$t = 61,060 \text{ s}$

Convert to hours:

$t = 61,060 \text{ s} \times \frac{1 \text{ min}}{60 \text{ s}} \times \frac{1 \text{ hour}}{60 \text{ min}}$

$t = \frac{61,060}{3600} \text{ hours}$

$t = 16.96 \text{ hours} \approx 17 \text{ hours}$

Answer (b):

$\boxed{t = 6.1 \times 10^4 \text{ s} \text{ or } 17 \text{ hours}}$

Interpretation:

• To drop from 0.30 M to 0.10 M takes 17 hours
• Concentration reduced to 1/3 of original
• Much longer than 1 hour (part a)
• Shows exponential decay nature

(c) Half-life

For first-order reactions, half-life is constant:

$t_{1/2} = \frac{0.693}{k} = \frac{\ln 2}{k}$

Substitute k:

$t_{1/2} = \frac{0.693}{1.8 \times 10^{-5}}$

$t_{1/2} = 3.85 \times 10^4 \text{ s}$

Convert to hours:

$t_{1/2} = \frac{3.85 \times 10^4}{3600} \text{ hours}$

$t_{1/2} = 10.7 \text{ hours}$

Answer (c):

$\boxed{t_{1/2} = 3.85 \times 10^4 \text{ s} \text{ or } 10.7 \text{ hours}}$

Verification using half-life:

After one half-life (10.7 hours):

• [H₂O₂] drops from 0.30 M to 0.15 M

After two half-lives (21.4 hours):

• [H₂O₂] drops from 0.15 M to 0.075 M

To go from 0.30 M to 0.10 M:

$\frac{0.30}{0.10} = 3 = 2^{1.585}$

So time needed:

$t = 1.585 \times t_{1/2} = 1.585 \times 10.7 = 17.0 \text{ hours}$

Matches part (b)! ✓

Summary Table

Key Concepts Illustrated

1. First-order characteristics:

Constant half-life:

• t₁/₂ = 10.7 hours (independent of concentration)
• After each 10.7 hours, concentration halves
• Exponential decay pattern

Linear ln[A] vs t:

• If plotted, would give straight line
• Slope = -k
• Intercept = ln[A]₀

2. Time conversions:

Always check units!

• k given in s⁻¹
• Time must be in seconds
• Convert hours → seconds for calculations
• Can convert back to hours for answer

3. Integrated rate law applications:

Forward calculation (given t, find [A]):

• Use: ln[A]_t = ln[A]₀ - kt
• Or: [A]_t = [A]₀e^(-kt)

Reverse calculation (given [A], find t):

• Rearrange: t = (1/k)ln([A]₀/[A]_t)

Half-life:

• Special case: [A]_t = [A]₀/2
• Gives: t₁/₂ = ln(2)/k = 0.693/k

Real-world context:

H₂O₂ decomposition:

Storage:

• Brown bottles (blocks light)
• Light catalyzes decomposition
• Cool, dark place preferred

Catalysts:

• MnO₂: very fast decomposition
• Enzymes (catalase): extremely fast
• Used to demonstrate catalysis

Applications:

• Bleaching agent
• Disinfectant
• Rocket propellant (concentrated)
• Must monitor concentration over time

Medical use:

• 3% solution (drugstore)
• Slow decomposition allows storage
• Eventually loses potency (O₂ escapes)

Graphical representation:

Concentration vs time:

• Exponential decay curve
• Starts at 0.30 M
• Asymptotically approaches 0
• After 10.7 h: 0.15 M
• After 21.4 h: 0.075 M

ln[H₂O₂] vs time:

• Straight line
• Slope = -k = -1.8 × 10⁻⁵ s⁻¹
• Useful for determining if reaction is first order

Practice tips:

For first-order problems:

1. ✓ Identify it's first order (given or determined)
2. ✓ Use ln[A]_t = ln[A]₀ - kt
3. ✓ Check time units match k units
4. ✓ For half-life: t₁/₂ = 0.693/k (constant!)
5. ✓ Remember ln(a/b) = ln(a) - ln(b)
6. ✓ Calculator: ln is natural log (base e)

Common mistakes:

• ✗ Using log instead of ln
• ✗ Mixing time units (hours vs seconds)
• ✗ Forgetting negative sign in rate law
• ✗ Using wrong integrated rate law for order