The rate of a certain chemical reaction is directly proportional to the square of the concentration of chemical A present and inversely proportional to the concentration of chemical B present. If the concentration of chemical B is increased by 100 percent, which of the following is closest to the percent change in the concentration of chemical A required to keep the reaction rate unchanged?
A. 100% decrease
B. 50% decrease
C. 40% decrease
D. 40% increase
E. 50% increase
Solution Section:
- 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
- Set up the proportional relationship: Express the reaction rate in terms of concentrations A and B using the given direct and inverse proportional relationships
- Apply the constraint of constant reaction rate: Create an equation showing the relationship between original and new concentrations when the rate remains unchanged
- Calculate the required change in concentration A: Solve for the new concentration of A and determine the percent change needed
Execution of Strategic Approach:
- 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
- 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 = k Ć (AĀ²/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: A_new = ?, B_new = 2B (since B increased by 100%)
- Apply the constraint of constant reaction rate Since we want the reaction rate to stay unchanged, we need: Original Rate = New Rate
This means: k Ć (AĀ²/B) = k Ć (A_newĀ²/B_new)
We can cancel out k from both sides: AĀ²/B = A_newĀ²/B_new
Now substituting B_new = 2B: AĀ²/B = A_newĀ²/(2B)
Multiplying both sides by B: AĀ² = A_newĀ²/2
Multiplying both sides by 2: 2AĀ² = A_newĀ²
- Calculate the required change in concentration A From 2AĀ² = A_newĀ², we can solve for A_new: A_newĀ² = 2AĀ²
Taking the square root of both sides: A_new = Aā2
Since ā2 ā 1.41, we have: A_new ā 1.41A
This means the new concentration is about 1.41 times the original concentration.
The percent change is: (A_new – A)/A Ć 100% = (1.41A – A)/A Ć 100% = 0.41A/A Ć 100% = 41%
Since A_new > 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 AĀ². 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/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., B_new = B) rather than understanding it means the value doubles (i.e., B_new = 2B).
Errors while executing the approach Faltering Point 1: Algebraic manipulation errors when solving 2AĀ² = A_newĀ². Students might incorrectly take the square root and get A_new = Aā2/2 instead of A_new = Aā2, essentially putting the 2 in the wrong place.
Faltering Point 2: Approximation errors with ā2. Students might use incorrect approximations like ā2 ā 1.5 instead of ā2 ā 1.41, leading to a 50% increase instead of 40%.
Faltering Point 3: Calculation errors when computing the percentage change formula (A_new – A)/A Ć 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 R = k Ć AĀ²/B, where k is a constant. Let’s choose smart numbers that make calculations clean.
Initial conditions: Let A = 10 and B = 5, so k = 1 for simplicity Initial rate: R = 1 Ć 10Ā²/5 = 100/5 = 20
Step 2: Apply the given change B increases by 100%, so new B = 5 + (100% of 5) = 5 + 5 = 10 We need the rate to remain 20, so: 20 = 1 Ć (new A)Ā²/10
Step 3: Solve for the new concentration of A 20 = (new A)Ā²/10 200 = (new A)Ā² new A = ā200 = ā(100 Ć 2) = 10ā2 ā 14.14
Step 4: Calculate the percent change Change in A = 14.14 – 10 = 4.14 Percent change = (4.14/10) Ć 100% = 41.4% ā 40% increase
Verification: New rate = 1 Ć (14.14)Ā²/10 = 200/10 = 20 ā
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.
