Problem 6 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

Equal quantities of gaseous hydrogen and iodine are mixed resulting in the reaction $$ \mathrm{H}_{2}+\mathrm{I}_{2} \longrightarrow 2 \mathrm{HI} $$ which runs until \(I_{2}\) is exhausted \(\left(H_{2}\right.\) is also exhausted). The rate at which \(I_{2}\) disappears is \(\frac{0.2}{(t+1)^{2}}\) \(\mathrm{gm} / \mathrm{sec} .\) How much \(I_{2}\) was initially introduced into the mixture? a. Sketch the graph of the reaction rate, \(r(t)=\frac{0.2}{(t+1)^{2}}\). b. Approximately how much \(I_{2}\) combined with \(H_{2}\) during the first second? c. Approximately how much \(I_{2}\) combined with \(H_{2}\) during the second second? d. Let \(Q(x)\) be the amount of \(I_{2}\) that combines with \(H_{2}\) during time 0 to \(x\) seconds. Write an integral that is \(Q(x)\) e. What is \(Q^{\prime}(x) ?\) f. Compute \(W^{\prime}(x)\) for \(W(x)=\frac{-0.2}{1+x}\). g. Show that there is a number, \(C,\) for which \(Q(x)=W(x)+C\). h. Show that \(C=0.2\) so that \(Q(x)=0.2-\frac{0.2}{1+x}\). i. How much \(I_{2}\) combined with \(H_{2}\) during the first second? j. How much \(I_{2}\) combined with \(H_{2}\) during the first 100 seconds? k. How much \(I_{2}\) combined with \(H_{2} ?\)

Step-by-Step Solution

Answer

The initial amount of \( I_2 \) was 0.2 grams.

Step 1: Graph of the Reaction Rate

The reaction rate is given as \( r(t) = \frac{0.2}{(t+1)^2} \). To sketch this graph, note that as \( t \) increases, the denominator \( (t+1)^2 \) increases, making \( r(t) \) decrease. Thus, the graph starts with a higher rate at \( t = 0 \) and asymptotically approaches zero as \( t \to \infty \).

Step 2: Amount of I2 Combined in the First Second

To find how much \( I_2 \) combined during the first second, evaluate the integral \( \int_{0}^{1} \frac{0.2}{(t+1)^2} \, dt \). This integral calculates the change in \( I_2 \) amount from time 0 to 1 second.

Step 3: Amount of I2 Combined in the Second Second

Similarly, to find the amount of \( I_2 \) combined in the second second, calculate \( \int_{1}^{2} \frac{0.2}{(t+1)^2} \, dt \). This integral represents the change in \( I_2 \) from time 1 to 2 seconds.

Step 4: Integral Expression for Q(x)

To write \( Q(x) \), the amount of \( I_2 \) that combines with \( H_2 \) from time 0 to \( x \), consider the integral \( Q(x) = \int_{0}^{x} \frac{0.2}{(t+1)^2} \, dt \).

Step 5: Derivative of Q(x)

By the Fundamental Theorem of Calculus, the derivative \( Q'(x) = \frac{0.2}{(x+1)^2} \). This confirms that integrating the rate function over time gives \( Q(x) \).

Step 6: Derivative of W(x)

For \( W(x) = \frac{-0.2}{1+x} \), compute \( W'(x) = \frac{0.2}{(1+x)^2} \). This shows \( W(x) \) has the same form as the negative integral result.

Step 7: Relationship Between Q(x) and W(x)

Use the fact that \( Q'(x) = -W'(x) \) to hypothesize \( Q(x) = W(x) + C \). Solve for \( C \) using known initial conditions.

Step 8: Determine Constant C

At \( t = 0 \), \( Q(0) = 0 \) since no \( I_2 \) has reacted. As \( W(0) = \frac{-0.2}{1+0} = -0.2 \), solve \( 0 = -0.2 + C \) to find \( C = 0.2 \).

Step 9: Final expression for Q(x)

Substitute \( C = 0.2 \) in \( Q(x) = W(x) + C \) to obtain \( Q(x) = 0.2 - \frac{0.2}{1+x} \).

Step 10: Amount of I2 Combined in the First Second

Substitute \( x = 1 \) in \( Q(x) = 0.2 - \frac{0.2}{1+x} \) to find \( Q(1) = 0.2 - \frac{0.2}{2} = 0.1 \).

Step 11: Amount of I2 Combined in the First 100 Seconds

Substitute \( x = 100 \) in \( Q(x) = 0.2 - \frac{0.2}{1+x} \) to calculate \( Q(100) = 0.2 - \frac{0.2}{101} \approx 0.198 \).

Step 12: Total Amount of I2 Combined

Evaluate \( Q(\infty) = 0.2 - \lim_{x \to \infty} \frac{0.2}{1+x} = 0.2 \), indicating all \( 0.2 \) grams initially present.

Key Concepts

Chemical Reaction Kinetics

Chemical reaction kinetics deals with the speed or rate at which chemical reactions occur. It focuses on understanding how different variables affect the reaction rate, such as:

• Temperature: Increasing temperature typically increases reaction rates because particles move faster and collide more frequently.
• Concentration: Higher concentrations usually lead to higher reaction rates, as there are more particles available to react.
• Catalysts: These substances can speed up a reaction without being consumed in the process.

In our exercise, the reaction between hydrogen and iodine produces hydrogen iodide, and we're observing the rate at which iodine, denoted by the formula \( r(t) = \frac{0.2}{(t+1)^2} \), disappears. This expression indicates that the rate depends inversely on the square of time, \( (t+1)^2 \), meaning the rate slows down as time proceeds. Initially, at \( t = 0 \), the rate is highest and decreases over time.

Differential Equations in Chemistry

Differential equations are equations involving derivatives, which can describe rates of change. In chemistry, they are often used to model reaction rates and phenomena where quantities change over time. For our problem, we have several steps concerning differential equations:

• The task involves integrating the rate function \( \frac{0.2}{(t+1)^2} \) with respect to time \( t \). This gives us \( Q(x) = \int_{0}^{x} \frac{0.2}{(t+1)^2} \, dt \), representing the cumulative amount of iodine that has reacted up to time \( x \).
• The derivative of this integral, \( Q'(x) = \frac{0.2}{(x+1)^2} \), reconfirms it returns the initial rate function, maintaining the reaction kinetics consistent over time intervals.
• The exercise solution involves calculating explicit values through integration over specific time intervals, which helps us understand the quantities of reactants transformed over time.

Differential equations in this context provide a framework to quantitatively analyze how much reactant transforms as time elapses, which is crucial in predicting outcomes of chemical processes.

Graphical Analysis of Functions

Graphical analysis is a powerful tool that helps in visualizing changes and behaviors of functions over time. In our scenario, creating a graph of the reaction rate function \( r(t) = \frac{0.2}{(t+1)^2} \) allows us to observe key characteristics:

• Initially, the function has a steeper slope, indicating a high reaction rate, which diminishes as \( t \) increases. This reflects how rapidly the iodine is consumed at the start.
• As \( t \) approaches infinity, \( r(t) \rightarrow 0 \), showing how the reaction rate approaches zero, indicating the reaction's completion.
• Using integral calculus, sequences like \( Q(x) = 0.2 - \frac{0.2}{1+x} \) show cumulative changes in iodine quantity over time. This complements the graph by providing specific amounts at given times, enhancing understanding of the process dynamics.

Such analysis isn't just about plotting curves but interpreting them to make sense of the physical phenomena, thus bridging theoretical results with practical interpretations.