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Joined: 20 Feb 2017
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The rate of a certain chemical reaction is directly proportional to th
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23 Nov 2021, 07:13
23 Nov 2021, 07:13
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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
(A) 100% decrease
(B) 50% decrease
(C) 40% decrease
(D) 40% increase
(E) 50% increase
The rate of a certain chemical reaction is directly proportional to th
[#permalink]
23 Nov 2021, 08:17
23 Nov 2021, 08:17
1
Rate, \(r ∝ A^2\) and \(r ∝ \frac{1}{B}\)
Therefore, \(r ∝ \frac{A^2}{B} \Rightarrow r = k\frac{A^2}{B}\), where \(k\) is a constant
\(B\) increases by 100%, therefore, \(B_{new} = B + 100\%B = 2B\)
Rate, \(r\) remains unchanged
Therefore, \(\frac{{A_{new}}^2}{B_{new}} = \frac{A^2}{B}\)
\(\frac{{A_{new}}^2}{2B} = \frac{A^2}{B}\)
\(\frac{{A_{new}}^2}{2} = \frac{A^2}{1}\)
\({A_{new}}^2 = 2A^2 \Rightarrow A_{new} = \sqrt{2A^2} = \sqrt{2}A\)
Percentage change in \(A = \frac{A_{new} - A}{A}*100\% = \frac{\sqrt{2}A - A}{A}*100\%\)
Note that \(\sqrt{2} ≈ 1.4142...\)
\(\frac{\sqrt{2}A - A}{A}*100\% \)
\(= \frac{1.414 - 1}{1}*100\% \)
\(= +41.42\% ≈ \textbf{+40%}\)
Therefore, 40% increase
Hence, Answer is D
Therefore, \(r ∝ \frac{A^2}{B} \Rightarrow r = k\frac{A^2}{B}\), where \(k\) is a constant
\(B\) increases by 100%, therefore, \(B_{new} = B + 100\%B = 2B\)
Rate, \(r\) remains unchanged
Therefore, \(\frac{{A_{new}}^2}{B_{new}} = \frac{A^2}{B}\)
\(\frac{{A_{new}}^2}{2B} = \frac{A^2}{B}\)
\(\frac{{A_{new}}^2}{2} = \frac{A^2}{1}\)
\({A_{new}}^2 = 2A^2 \Rightarrow A_{new} = \sqrt{2A^2} = \sqrt{2}A\)
Percentage change in \(A = \frac{A_{new} - A}{A}*100\% = \frac{\sqrt{2}A - A}{A}*100\%\)
Note that \(\sqrt{2} ≈ 1.4142...\)
\(\frac{\sqrt{2}A - A}{A}*100\% \)
\(= \frac{1.414 - 1}{1}*100\% \)
\(= +41.42\% ≈ \textbf{+40%}\)
Therefore, 40% increase
Hence, Answer is D
gmatclubot
The rate of a certain chemical reaction is directly proportional to th [#permalink]
23 Nov 2021, 08:17
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