Problem 17 - Essential Calculus

Question

In an autocatalytic chemical reaction, the product formed is a catalyst for the reaction. If \(Q_{0}\) is the amount of the original substance and \(x\) is the amount of catalyst formed, the rate of chemical reaction is \(\frac{d Q}{d x}=k x\left(Q_{0}-x\right)\) For what value of \(x\) will the rate of chemical reaction be greatest?

Step-by-Step Solution

Answer

The rate of the chemical reaction will be greatest when \(x = \frac{Q_{0}}{2}\)

Step 1: Identify the function to be maximized

The function describing the rate of the chemical reaction is \(\frac{d Q}{d x}=k x\left(Q_{0}-x\right)\)

Step 2: Differentiate the function

Differentiate the function with respect to \(x\), to obtain: \(\frac{d^{2} Q}{d x^{2}}=k\left(Q_{0}-2 x\right)\)

Step 3: Determine the maximum point

The maximum rate of the reaction occurs when \(\frac{d^{2} Q}{d x^{2}} = k\left(Q_{0}-2 x\right) = 0\). Solving this equation for \(x\) yields \(x = \frac{Q_{0}}{2}\)