Problem 11 - Advanced Engineering Mathematics

Question

Two chemicals \(A\) and \(B\) are combined to form a chemical \(C\). The rate, or velocity, of the reaction is proportional to the product of the instantaneous amounts of \(A\) and \(B\) not converted to chemical \(C\). Initially there are 40 grams of \(A\) and 50 grams of \(B\), and for each gram of \(B\), 2 grams of \(A\) is used. It is observed that 10 grams of \(C\) is formed in 5 minutes. How much is formed in 20 minutes? What is the limiting amount of \(C\) after a long time? How much of chemicals \(A\) and \(B\) remains after a long time?

Step-by-Step Solution

Answer

20 grams of \(C\) is eventually formed. No \(A\) remains, and 30 grams of \(B\) are left.

Step 1: Set Up the Reaction Rate Equation

The rate of reaction (velocity) can be expressed as \( \frac{dC}{dt} = k imes (a - C) \times \left(b - \frac{C}{2}\right) \), where \(C\) is the amount of chemical \(C\) formed at time \(t\), \(a\) and \(b\) are the initial amounts of chemicals \(A\) and \(B\) respectively, and \(k\) is the rate constant. Given \(a = 40\) grams and \(b = 50\) grams.

Step 2: Calculate the Rate Constant \(k\)

Given that 10 grams of \(C\) is formed in 5 minutes, we can substitute this into our rate equation to find \(k\). Using the equation \(10 = k \times (40-10) \times \left(50-\frac{10}{2}\right) \times 5\), solve for \(k\). This simplifies to \(10 = 5k(30)(45)\), therefore, \(k = \frac{1}{675}\).

Step 3: Calculate the Amount of \(C\) Formed in 20 Minutes

Given \(k = \frac{1}{675}\), use the same rate equation to find \(C\) after 20 minutes: \(C = k \times (40-C) \times \left(50-\frac{C}{2}\right) \times 20\). This requires solving a differential equation, or approximating based on the growth pattern observed at 5 minutes. Assume polynomial growth or exponential decay as an approximation to solve.

Step 4: Determine the Limiting Amount of \(C\)

In the long term, \(A\) and \(B\) will be fully converted based on stoichiometry. Since 2 grams of \(A\) combine with 1 gram of \(B\) to form \(C\), the reaction is limited by the amount of \(A\). Thus, \(A = 40\) grams and \(B = 50\) grams imply that \(C_{\text{limiting}} = 40 / 2 = 20 \) grams, \(B_{\text{excess}} = 50 - 20 = 30 \) grams remain.

Step 5: Calculate Remaining \(A\) and \(B\) After a Long Time

Since all of \(A\) is converted, no \(A\) remains. The excess amount of \(B\) is left over since \(b - \frac{C}{2} = 50 - \frac{20}{2} = 40 - 20 = 30 \) grams remain.

Key Concepts

reaction rate

In chemical kinetics, the reaction rate is a measure of how quickly a chemical reaction occurs. It is often defined as the change in concentration of reactants or products over time. For the reaction between chemicals \(A\) and \(B\) to form \(C\), the rate can be mathematically expressed using the formula:

• \( \frac{dC}{dt} = k \times (a - C) \times \left(b - \frac{C}{2}\right) \)

Here, \(C\) represents the amount of chemical \(C\) formed at time \(t\). The variables \(a\) and \(b\) represent the initial quantities of chemicals \(A\) and \(B\) respectively, while \(k\) is the reaction rate constant.
This equation illustrates that the reaction rate depends on the amount of \(A\) and \(B\) available and not yet converted to \(C\).
Understanding reaction rate is crucial. It helps predict how quickly a reaction will proceed and how much product will form over a given time.

stoichiometry

Stoichiometry involves the calculation of reactants and products in chemical reactions. It is grounded on the conservation of mass and chemical formulas. In our scenario, stoichiometry dictates how chemical \(C\) is formed from \(A\) and \(B\).

• For every 2 grams of \(A\), 1 gram of \(B\) is used to form chemical \(C\).

This balanced proportion is essential as it establishes the theoretical limits of the reaction.
The reaction can only continue as long as the reactants are available in these exact ratios. It helps determine what the 'limiting reactant' is, and consequently, the maximum amount of product that can be formed.
In this problem, \(A\) is the limiting reactant because it runs out first when combined with \(B\), resulting in a maximum of 20 grams of \(C\) formed.

differential equations

Differential equations are powerful mathematical tools used to describe change. They are employed in chemical kinetics to model the rate of reaction as a function of the concentration of reactants over time.

• The reaction rate equation: \( \frac{dC}{dt} = k \times (a - C) \times \left(b - \frac{C}{2}\right) \) is a first-order differential equation.

Solving this equation gives the quantity of \(C\) formed at any time \(t\) during the reaction.
This involves integration techniques, but it also highlights the dynamic nature of chemical reactions.
For students, understanding such equations can seem complex, but breaking it down shows how concentrations of chemicals change at each moment of the process. This is vital for predicting future concentrations and for controlling reactions in industrial applications.