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 teh 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
Would someone mind explaining how they solved this problem? OA is D
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Rate of Chemical Reaction
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GMATGuruNY
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x is directly proportional to y, translated into math: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 precent 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
OA is D
x = ky, where k is a constant.
x is inversely proportional to z, translated into math:
x = k/z, where is k is a constant.
x is directly proportional to y and inversely proportional to z, translated into math:
x = k(y/z), where k is a constant.
In the problem above, the rate is directly proportional to the square of A and inversely proportional to B.
Thus:
R = k(A²/B).
Original values:
Let:
A = 10
B = 1
k = 1.
Then:
R = (1)(10²/1)
R = 100.
New values:
Since the rate doesn't change, R = 100.
B increased by 100% = 2.
k = 1. (Since k is a constant.)
Solving for A, we get:
100 = (1)(A²/2)
A² = 200
A = √200 = √100 * √2 = 10√2 ≈ 10(1.4) = 14.
Percent increase in A:
(increase in A)/(original A) = (14-10)/10 = 4/10 = 40%.
The correct answer is D.
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We can let n = the rate of a certain chemical reaction, a = the concentration of chemical A, and b = the concentration of chemical B. We are given that 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, so, for some positive constant k, we have:vrn2vw wrote: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 teh 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
Would someone mind explaining how they solved this problem? OA is D
n = ka^2/b
When b is increased by 100 percent, b becomes 2b. To keep the reaction rate unchanged, we can let a become c, so we have:
ka^2/b = kc^2/(2b)
2bka^2 = bkc^2
2a^2 = c^2
c = √(2a^2)
c = a√2
Since √2 ≈ 1.4, c ≈ 1.4a or approximately 140% of a, i.e., a 40% increase in the concentration of chemical A.
Answer: D
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