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

Thanks =) do you have any links to similar questions? i'm very shaky on these

This question can also be solved by plugging in values for a and b and comparing the results:

- The rate of this reaction at first is \(r=\frac{A^2}{B}\)
- If \(A=1\) and \(B=2\) the result of the equation is \(r=\frac{1}{2}\)
- When \(B\) is increased by 100% the equation becomes \(r=\frac{A^2}{2B}\)
- Since the question asks by how much \(A\) would need to increase in order for the reaction rate to remain the same, the second equation can be rewritten as \(\frac{1}{2}=\frac{A^2}{4}\) where \(r=\frac{1}{2}\), the same as in the first equation.
- The second equation can then be rewritten as \(2=A^2\) and then as \(\sqrt{2}=A\)
- The percent increase from \(A=1\) in the first equation to \(A=\sqrt{2}\) in the second equation is approximately 40% since\(\sqrt{2}=1.4\)
- The answer is
Spoiler: ::
D

\(rate = {\text{cte}} \cdot \frac{{{{\left[ A \right]}^2}}}{{\left[ B \right]}}\,\,\,\,\,\left( {{\text{cte}} \ne 0} \right)\,\,\,\,\,\,\mathop \Rightarrow \limits^{{\text{rate}}\,\,{\text{unchanged}}} \,\,\,\,\,{\text{cte}} \cdot \frac{{{{\left[ {{A_{\,\,2}}} \right]}^2}}}{{\left[ {{B_{\,2}}} \right]}} = {\text{cte}} \cdot \frac{{{{\left[ {{A_{\,\,1}}} \right]}^2}}}{{\left[ {{B_{\,1}}} \right]}}\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{{\text{cte}}\,\, \ne \,\,{\text{0}}} \,\,\,\,\,\,{\left( {\frac{{\left[ {{A_{\,2}}} \right]}}{{\left[ {{A_{\,1}}} \right]}}} \right)^{\,2}} = \frac{{\left[ {{B_{\,2}}} \right]}}{{\left[ {{B_{\,1}}} \right]}}\)

\(\left[ {{B_{\,2}}} \right] = \left[ {2{B_{\,1}}} \right]\,\,\,\,\,\,;\,\,\,\,\,\,\left[ {{A_{\,2}}} \right] = \left[ {k{A_{\,1}}} \right]\,\,\,\,\,\,\,\left( {k > 0} \right)\)


\(?\,\, \cong \,\,k - 1\)


\({\left( {\frac{{\left[ {{A_{\,2}}} \right]}}{{\left[ {{A_{\,1}}} \right]}}} \right)^{\,2}} = \frac{{\left[ {{B_{\,2}}} \right]}}{{\left[ {{B_{\,1}}} \right]}}\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\left\{ \begin{gathered}
{k^2} = 2 \hfill \\
\sqrt 2 \cong 1.41 \hfill \\
\end{gathered} \right.\,\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{k\,\, > \,\,0} \,\,\,\,\,\,?\,\, = \,\,\sqrt 2 - 1\,\,\, \cong \,\,\,0.41\,\,\, = \,\,\,41\% \,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\sqrt 2 - 1\,\, > 0\,\,\,\, \Rightarrow \,\,\,{\text{increase}}} \right)\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Given: 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.

Asked: 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?

Rate of chemical reaction = constant * (concentration of chemical A)^2 / (concentration of chemical B)
r = k * Ca^2 / Cb

Cb' = 2Cb

r = k Ca'^2/2Cb

Ca'^2 = 2Ca^2
Ca' = \sqrt 2 * Ca
Ca' = 1.4 * Ca
40% increase in concentration of chemical A

IMO D
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Given:
r = (k.(Ca)^2)/Cb (where Ca is conc. of chemical A and Cb is conc. of chemical B)
Cb increases by 100% => Cb' = 2Cb (Cb'is the new conc. of chemical B)
r is same =>
(Ca)^2 / Cb = (Ca')^2/Cb' (Ca'is the new conc. of chemical A)
Cb' = 2Cb
(Ca')^2 = 2(Ca)^2 => Ca' = 1.41Ca (Implying 40% increase in existing conc. of chemical A)

Ans: D
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