A chemical reaction has a rate given by the formula \( R = k[A]^2[B] \). If the concentration of \( A \) is doubled and the concentration of \( B \) is halved, how does the reaction rate \( R \) change?

["How Reaction Rate Changes When [A] Is Doubled and [B] Is Halved – Understanding Rate Laws with the Formula ( R = k[A]^2[B] )", "Understanding the factors that influence chemical reaction rates is essential in chemistry and chemical engineering. The rate of a reaction, expressed by the rate law ( R = k[A]^2[B] ), depends on the concentrations of reactants and the rate constant ( k ). This article explores how changing the concentrations of reactants ( A ) and ( B ) affects the overall reaction rate, particularly when ( [A] ) is doubled and ( [B] ) is halved.", "---", "### The Rate Law: ( R = k[A]^2[B] )", "The rate law clearly shows that:", "- The reaction is second-order with respect to ( A ), meaning the rate depends on the square of ( [A] ).\n- The reaction is first-order with respect to ( B ), so the rate is directly proportional to ( [B] ).", "---", "### Applying the Concentration Changes", "Let’s analyze the effect of altering the concentrations:", "- Original rate: ( R = k[A]^2[B] )\n- New concentration of ( A ): ( [A]' = 2[A] )\n- New concentration of ( B ): ( [B]' = \frac{1}{2}[B] )", "Now substitute these into the rate law:", "[\nR' = k[A']^2[B'] = k(2[A])^2\left(\frac{1}{2}[B]\right)\n]", "Simplify step-by-step:", "[\nR' = k \cdot 4[A]^2 \cdot \frac{1}{2}[B] = k \cdot 2[A]^2[B]\n]", "Thus, ( R' = 2k[A]^2[B] = 2R )", "---", "### Conclusion: The Reaction Rate Doubles", "When the concentration of ( A ) is doubled and the concentration of ( B ) is halved, the reaction rate ( R ) doubles. This illustrates the power of reaction order—especially the squared dependence on ( [A] )—in determining how concentration changes impact the overall rate.", "---", "### Why This Matters", "Understanding these relationships helps scientists and engineers predict reaction behavior under different conditions. It supports process optimization in industries ranging from pharmaceuticals to petrochemicals, where controlling reaction speeds ensures efficiency, safety, and product quality.", "[\n\boxed{\ ext{Doubling } [A] \ ext{ and halving } [B] \ ext{ results in a twofold increase in the reaction rate } R.}\n]"]