The rate of a certain chemical reaction is directly proportional to the square of the concentration of chemical A present...

1. Translate the problem requirements: Convert the proportional relationship into a mathematical equation and clarify what "keep the reaction rate unchanged" means when one variable changes
2. Set up the proportional relationship: Express the reaction rate in terms of concentrations A and B using the given direct and inverse proportional relationships
3. Apply the constraint of constant reaction rate: Create an equation showing the relationship between original and new concentrations when the rate remains unchanged
4. Calculate the required change in concentration A: Solve for the new concentration of A and determine the percent change needed

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what this problem is telling us in plain English.

We have a chemical reaction whose speed depends on two chemicals:

• Chemical A: The MORE of chemical A we have, the FASTER the reaction goes. But it's not just a simple relationship - if we double chemical A, the reaction goes 4 times faster (because it depends on the square of A's concentration)
• Chemical B: The MORE of chemical B we have, the SLOWER the reaction goes (inverse relationship)

The problem tells us that chemical B's concentration increases by 100% (meaning it doubles), and we need to find how much chemical A's concentration must change to keep the reaction speed exactly the same.

Process Skill: TRANSLATE - Converting the proportional language into clear mathematical understanding

2. Set up the proportional relationship

Now let's express this mathematically. Since the reaction rate is:

• Directly proportional to the square of A's concentration
• Inversely proportional to B's concentration

We can write: Reaction Rate = \(\mathrm{k} \times (\mathrm{A}^2/\mathrm{B})\), where k is some constant.

Let's use simple numbers to make this concrete:

• Original concentrations: A = some value, B = some value
• New concentrations: \(\mathrm{A_{new}} = ?\), \(\mathrm{B_{new}} = 2\mathrm{B}\) (since B increased by 100%)

3. Apply the constraint of constant reaction rate

Since we want the reaction rate to stay unchanged, we need:

Original Rate = New Rate

This means: \(\mathrm{k} \times (\mathrm{A}^2/\mathrm{B}) = \mathrm{k} \times (\mathrm{A_{new}}^2/\mathrm{B_{new}})\)

We can cancel out k from both sides: \(\mathrm{A}^2/\mathrm{B} = \mathrm{A_{new}}^2/\mathrm{B_{new}}\)

Now substituting \(\mathrm{B_{new}} = 2\mathrm{B}\):

\(\mathrm{A}^2/\mathrm{B} = \mathrm{A_{new}}^2/(2\mathrm{B})\)

Multiplying both sides by B: \(\mathrm{A}^2 = \mathrm{A_{new}}^2/2\)

Multiplying both sides by 2: \(2\mathrm{A}^2 = \mathrm{A_{new}}^2\)

4. Calculate the required change in concentration A

From \(2\mathrm{A}^2 = \mathrm{A_{new}}^2\), we can solve for \(\mathrm{A_{new}}\):

\(\mathrm{A_{new}}^2 = 2\mathrm{A}^2\)

Taking the square root of both sides: \(\mathrm{A_{new}} = \mathrm{A}\sqrt{2}\)

Since \(\sqrt{2} \approx 1.41\), we have: \(\mathrm{A_{new}} \approx 1.41\mathrm{A}\)

This means the new concentration is about 1.41 times the original concentration.

The percent change is: \((\mathrm{A_{new}} - \mathrm{A})/\mathrm{A} \times 100\% = (1.41\mathrm{A} - \mathrm{A})/\mathrm{A} \times 100\% = 0.41\mathrm{A}/\mathrm{A} \times 100\% = 41\%\)

Since \(\mathrm{A_{new}} > \mathrm{A}\), this is an increase of approximately 40%.

Final Answer

The concentration of chemical A must increase by approximately 40% to keep the reaction rate unchanged when chemical B's concentration doubles.

Answer: (D) 40% increase

Common Faltering Points

Errors while devising the approach

Faltering Point 1: Misunderstanding the proportional relationships. Students often confuse "directly proportional to the square of A" and write the rate as proportional to A instead of \(\mathrm{A}^2\). This fundamental error in setting up the basic relationship equation will lead to completely incorrect calculations.

Faltering Point 2: Incorrectly interpreting "inversely proportional to B." Some students might write the rate as proportional to B instead of \(1/\mathrm{B}\), missing the inverse relationship entirely.

Faltering Point 3: Misunderstanding what "increased by 100%" means. Students might think this means the new value is 100% of the original (i.e., \(\mathrm{B_{new}} = \mathrm{B}\)) rather than understanding it means the value doubles (i.e., \(\mathrm{B_{new}} = 2\mathrm{B}\)).

Errors while executing the approach

Faltering Point 1: Algebraic manipulation errors when solving \(2\mathrm{A}^2 = \mathrm{A_{new}}^2\). Students might incorrectly take the square root and get \(\mathrm{A_{new}} = \mathrm{A}\sqrt{2}/2\) instead of \(\mathrm{A_{new}} = \mathrm{A}\sqrt{2}\), essentially putting the 2 in the wrong place.

Faltering Point 2: Approximation errors with \(\sqrt{2}\). Students might use incorrect approximations like \(\sqrt{2} \approx 1.5\) instead of \(\sqrt{2} \approx 1.41\), leading to a 50% increase instead of 40%.

Faltering Point 3: Calculation errors when computing the percentage change formula \((\mathrm{A_{new}} - \mathrm{A})/\mathrm{A} \times 100\%\). Students might forget to subtract the original value A or make basic arithmetic mistakes in the final percentage calculation.

Errors while selecting the answer

Faltering Point 1: Confusing increase vs. decrease. If students made errors in their algebra, they might get a value less than the original concentration and incorrectly select a decrease option instead of an increase.

Alternate Solutions

Smart Numbers Approach

Step 1: Set up the relationship with concrete values

The reaction rate \(\mathrm{R} = \mathrm{k} \times \mathrm{A}^2/\mathrm{B}\), where k is a constant. Let's choose smart numbers that make calculations clean.

Initial conditions: Let \(\mathrm{A} = 10\) and \(\mathrm{B} = 5\), so \(\mathrm{k} = 1\) for simplicity

Initial rate: \(\mathrm{R} = 1 \times 10^2/5 = 100/5 = 20\)

Step 2: Apply the given change

B increases by 100%, so new \(\mathrm{B} = 5 + (100\% \text{ of } 5) = 5 + 5 = 10\)

We need the rate to remain 20, so: \(20 = 1 \times (\text{new A})^2/10\)

Step 3: Solve for the new concentration of A

\(20 = (\text{new A})^2/10\)

\(200 = (\text{new A})^2\)

\(\text{new A} = \sqrt{200} = \sqrt{100 \times 2} = 10\sqrt{2} \approx 14.14\)

Step 4: Calculate the percent change

Change in A = 14.14 - 10 = 4.14

Percent change = \((4.14/10) \times 100\% = 41.4\% \approx 40\%\) increase

Verification: New rate = \(1 \times (14.14)^2/10 = 200/10 = 20\) ✓

The smart numbers approach works well here because we can choose convenient initial values for A and B that make the arithmetic manageable while preserving the proportional relationships described in the problem.