Answer

Problem 1: Rate Law Determination

a. Order with respect to A and B

Order in B — compare Experiments 2 and 3, where [A] is held constant at 0.300 M:

$$\frac{\text{Rate}_3}{\text{Rate}_2} = \frac{0.150}{0.0375} = 4.00 \qquad \frac{[B]_3}{[B]_2} = \frac{0.200}{0.100} = 2.00$$

Since $2.00^n = 4.00$, $n = 2$. The reaction is second order in B.

Order in A — compare Experiments 1 and 4, where [B] is held constant at 0.0500 M:

$$\frac{\text{Rate}_4}{\text{Rate}_1} = \frac{0.00625}{0.00313} = 2.00 \qquad \frac{[A]_4}{[A]_1} = \frac{0.200}{0.100} = 2.00$$

Since $2.00^m = 2.00$, $m = 1$. The reaction is first order in A.

b. Rate law

$$\text{Rate} = k[A][B]^2$$

Overall order = 1 + 2 = 3rd order.

c. Calculating k

Solving $k = \dfrac{\text{Rate}}{[A][B]^2}$ for each trial:

Exp	Calculation	k
1	0.00313 / [(0.100)(0.0500)²]	12.5
2	0.0375 / [(0.300)(0.100)²]	12.5
3	0.150 / [(0.300)(0.200)²]	12.5
4	0.00625 / [(0.200)(0.0500)²]	12.5

k ≈ 12.5 M⁻² s⁻¹ (units follow from Rate/(M·M²) = M/s ÷ M³ = M⁻²s⁻¹)

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Problem 2: Flawed Dilution Design

a. Why it fails

The group changed the total volume between trials (4.00 mL → 6.00 mL) without adding a compensating diluent. Because the volume of H₂O₂ solution stayed the same (2.00 mL) but got diluted into a larger total volume in Trial 2, [H₂O₂] itself changes between trials even though the volume of H₂O₂ added didn’t change:

• Trial 1: [H₂O₂] = (0.88 M × 2.00 mL)/4.00 mL = 0.44 M
• Trial 2: [H₂O₂] = (0.88 M × 2.00 mL)/6.00 mL = 0.293 M

So two variables changed at once ([KI] doubled and [H₂O₂] dropped by a third), making it impossible to attribute any change in rate to [KI] alone.

b. Fix

Keep the total volume constant across all trials by adding water as a diluent to make up the difference. For example:

• Trial 1: 2.00 mL H₂O₂ + 2.00 mL KI + 2.00 mL H₂O = 6.00 mL total
• Trial 2: 2.00 mL H₂O₂ + 4.00 mL KI + 0.00 mL H₂O = 6.00 mL total

This holds [H₂O₂] fixed at (0.88 × 2.00)/6.00 = 0.293 M in both trials while cleanly doubling [KI] from 0.167 M to 0.333 M.

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Problem 3: Using Pressure Data to Find Rate (conceptual)

The KI-catalyzed decomposition of H₂O₂ produces O₂ gas:

$$2\text{H}_2\text{O}_2 \xrightarrow{\text{I}^-} 2\text{H}_2\text{O} + \text{O}_2(g)$$

If the reaction is run in a sealed flask connected to a pressure sensor, the O₂ produced has nowhere to go, so pressure rises over time as the reaction proceeds. Since pressure is directly proportional to moles of gas (ideal gas law, at fixed volume and temperature), the rate of pressure increase (slope of a pressure-vs-time plot, especially over the initial, roughly linear portion) is directly proportional to the rate of O₂ formation — and therefore, through the reaction stoichiometry, to the rate of H₂O₂ decomposition. You wouldn’t need to measure concentration directly at all; you’d track ΔP/Δt and convert it to a rate using PV = nRT (solving for moles of O₂ per unit time, then using stoichiometry to relate that to moles of H₂O₂ consumed per unit time, then dividing by the reaction volume to get M/s).

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Problem 4: Proposed Trial List

To determine the orders in both H₂O₂ and KI, vary one reactant at a time while holding the other constant, keeping total volume fixed at 10.00 mL using water as a diluent (stock solutions: 0.88 M H₂O₂, 0.50 M KI):

Trial	H₂O₂ (mL)	KI (mL)	H₂O (mL)	Total (mL)	[H₂O₂] (M)	[KI] (M)	Purpose
1	2.00	2.00	6.00	10.00	0.176	0.100	Baseline
2	4.00	2.00	4.00	10.00	0.352	0.100	2× [H₂O₂]
3	6.00	2.00	2.00	10.00	0.528	0.100	3× [H₂O₂]
4	2.00	4.00	4.00	10.00	0.176	0.200	2× [KI]
5	2.00	6.00	2.00	10.00	0.176	0.300	3× [KI]
6	2.00	2.00	6.00	10.00	0.176	0.100	Replicate of baseline (for reproducibility/std. dev.)

This design isolates each reactant’s effect on rate while keeping total volume — and therefore all other concentrations — constant, avoiding the flaw in Problem 2.

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Post-Lab Analysis (Method — depends on your actual collected data)

These last three items can only be completed with your group’s real pressure/time data, but here’s the method to apply once you have it:

1. Determine the rate law. For each trial, find the initial rate from the slope of the linear (early-time) region of your pressure-vs-time plot, converted to M/s via the ideal gas law and stoichiometry. Then compare trials pairwise exactly as in Problem 1: hold one concentration constant, compare how rate changes when the other doubles/triples, and solve for the exponent (order) in each reactant. Combine into Rate = k[H₂O₂]^m[KI]^n.

2. Calculate k. Once you have the orders, solve $k = \text{Rate}/([\text{H}_2\text{O}_2]^m[\text{KI}]^n)$ for each trial, then take the mean and sample standard deviation ($s = \sqrt{\frac{\sum(k_i – \bar{k})^2}{n-1}}$) across all trials (including your replicate).

3. Sources of uncertainty to consider discussing: temperature fluctuations affecting rate and gas volume (ideal gas law sensitivity), pressure sensor lag or leaks in the sealed system, imprecision in pipetting stock solutions, non-instantaneous mixing at t = 0 (affecting where “initial rate” is measured from), and any curvature in the pressure-vs-time plot that makes picking a truly “initial” linear slope subjective.