{"id":57724,"date":"2025-08-27T22:40:16","date_gmt":"2025-08-27T17:10:16","guid":{"rendered":"https:\/\/e-gmat.com\/blogs\/?p=57724"},"modified":"2025-09-04T23:04:40","modified_gmt":"2025-09-04T17:34:40","slug":"the-rate-of-a-certain-chemical-reaction-is-directly-proportional-to-the","status":"publish","type":"post","link":"https:\/\/e-gmat.com\/blogs\/the-rate-of-a-certain-chemical-reaction-is-directly-proportional-to-the\/","title":{"rendered":"The rate of a certain chemical reaction is directly proportional to the square of the….."},"content":{"rendered":"A <\/span> 4<\/span> min read <\/span><\/span>\n

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?<\/p>\n\n\n\n

A. 100% decrease<\/p>\n\n\n\n

B. 50% decrease<\/p>\n\n\n\n

C. 40% decrease<\/p>\n\n\n\n

D. 40% increase<\/p>\n\n\n\n

E. 50% increase<\/p>\n\n\n\n

Solution Section:<\/h2>\n\n\n\n
  • 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<\/li>
  • Set up the proportional relationship: Express the reaction rate in terms of concentrations A and B using the given direct and inverse proportional relationships<\/li>
  • Apply the constraint of constant reaction rate: Create an equation showing the relationship between original and new concentrations when the rate remains unchanged<\/li>
  • Calculate the required change in concentration A: Solve for the new concentration of A and determine the percent change needed<\/li><\/ul>\n\n\n\n

    Execution of Strategic Approach:<\/strong><\/h3>\n\n\n\n
    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:<\/li><\/ol>\n\n\n\n
      • 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)<\/li>
      • Chemical B: The MORE of chemical B we have, the SLOWER the reaction goes (inverse relationship)<\/li><\/ul>\n\n\n\n

        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.<\/p>\n\n\n\n

        Process Skill: TRANSLATE – Converting the proportional language into clear mathematical understanding<\/p>\n\n\n\n

        1. Set up the proportional relationship Now let’s express this mathematically. Since the reaction rate is:<\/li><\/ol>\n\n\n\n
          • Directly proportional to the square of A’s concentration<\/li>
          • Inversely proportional to B’s concentration<\/li><\/ul>\n\n\n\n

            We can write: Reaction Rate = k \u00d7 (A\u00b2\/B), where k is some constant.<\/p>\n\n\n\n

            Let’s use simple numbers to make this concrete:<\/p>\n\n\n\n

            • Original concentrations: A = some value, B = some value<\/li>
            • New concentrations: A_new = ?, B_new = 2B (since B increased by 100%)<\/li><\/ul>\n\n\n\n
              1. Apply the constraint of constant reaction rate Since we want the reaction rate to stay unchanged, we need: Original Rate = New Rate<\/li><\/ol>\n\n\n\n

                This means: k \u00d7 (A\u00b2\/B) = k \u00d7 (A_new\u00b2\/B_new)<\/p>\n\n\n\n

                We can cancel out k from both sides: A\u00b2\/B = A_new\u00b2\/B_new<\/p>\n\n\n\n

                Now substituting B_new = 2B: A\u00b2\/B = A_new\u00b2\/(2B)<\/p>\n\n\n\n

                Multiplying both sides by B: A\u00b2 = A_new\u00b2\/2<\/p>\n\n\n\n

                Multiplying both sides by 2: 2A\u00b2 = A_new\u00b2<\/p>\n\n\n\n

                1. Calculate the required change in concentration A From 2A\u00b2 = A_new\u00b2, we can solve for A_new: A_new\u00b2 = 2A\u00b2<\/li><\/ol>\n\n\n\n

                  Taking the square root of both sides: A_new = A\u221a2<\/p>\n\n\n\n

                  Since \u221a2 \u2248 1.41, we have: A_new \u2248 1.41A<\/p>\n\n\n\n

                  This means the new concentration is about 1.41 times the original concentration.<\/p>\n\n\n\n

                  The percent change is: (A_new – A)\/A \u00d7 100% = (1.41A – A)\/A \u00d7 100% = 0.41A\/A \u00d7 100% = 41%<\/p>\n\n\n\n

                  Since A_new > A, this is an increase of approximately 40%.<\/p>\n\n\n\n

                  Final Answer The concentration of chemical A must increase by approximately 40% to keep the reaction rate unchanged when chemical B’s concentration doubles.<\/p>\n\n\n\n

                  Answer: (D) 40% increase<\/p>\n\n\n\n

                  Common Faltering Points:<\/strong><\/h3>\n\n\n\n

                  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 A\u00b2. This fundamental error in setting up the basic relationship equation will lead to completely incorrect calculations.<\/p>\n\n\n\n

                  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.<\/p>\n\n\n\n

                  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).<\/p>\n\n\n\n

                  Errors while executing the approach Faltering Point 1: Algebraic manipulation errors when solving 2A\u00b2 = A_new\u00b2. Students might incorrectly take the square root and get A_new = A\u221a2\/2 instead of A_new = A\u221a2, essentially putting the 2 in the wrong place.<\/p>\n\n\n\n

                  Faltering Point 2: Approximation errors with \u221a2. Students might use incorrect approximations like \u221a2 \u2248 1.5 instead of \u221a2 \u2248 1.41, leading to a 50% increase instead of 40%.<\/p>\n\n\n\n

                  Faltering Point 3: Calculation errors when computing the percentage change formula (A_new – A)\/A \u00d7 100%. Students might forget to subtract the original value A or make basic arithmetic mistakes in the final percentage calculation.<\/p>\n\n\n\n

                  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.<\/p>\n\n\n\n

                  Alternate Solutions:<\/strong><\/h3>\n\n\n\n

                  Smart Numbers Approach<\/p>\n\n\n\n

                  Step 1: Set up the relationship with concrete values The reaction rate R = k \u00d7 A\u00b2\/B, where k is a constant. Let’s choose smart numbers that make calculations clean.<\/p>\n\n\n\n

                  Initial conditions: Let A = 10 and B = 5, so k = 1 for simplicity Initial rate: R = 1 \u00d7 10\u00b2\/5 = 100\/5 = 20<\/p>\n\n\n\n

                  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 \u00d7 (new A)\u00b2\/10<\/p>\n\n\n\n

                  Step 3: Solve for the new concentration of A 20 = (new A)\u00b2\/10 200 = (new A)\u00b2 new A = \u221a200 = \u221a(100 \u00d7 2) = 10\u221a2 \u2248 14.14<\/p>\n\n\n\n

                  Step 4: Calculate the percent change Change in A = 14.14 – 10 = 4.14 Percent change = (4.14\/10) \u00d7 100% = 41.4% \u2248 40% increase<\/p>\n\n\n\n

                  Verification: New rate = 1 \u00d7 (14.14)\u00b2\/10 = 200\/10 = 20 \u2713<\/p>\n\n\n\n

                  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.<\/p>\n","protected":false},"excerpt":{"rendered":"

                  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. 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