Problem 7 - Essential Calculus Early Transcendentals

Question

Experiments show that if the chemical reaction $$\mathrm{N}_{2} \mathrm{O}_{5} \rightarrow 2 \mathrm{NO}_{2}+\frac{1}{2} \mathrm{O}_{2}$$ takes place at \(45^{\circ} \mathrm{C},\) the rate of reaction of dinitrogen pentoxide is proportional to its concentration as follows: $$-\frac{d\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]}{d t}=0.0005\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]$$ (a) Find an expression for the concentration \(\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]\) after \(t\) seconds if the initial concentration is \(C .\) (b) How long will the reaction take to reduce the concen- tration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) to 90\(\%\) of its original value?

Step-by-Step Solution

Answer

(a) \([\mathrm{N}_2 \mathrm{O}_5] = C e^{-0.0005t}\); (b) Approximately 211 seconds.

Step 1: Set Up the Differential Equation

We are given a rate of reaction equation: \(-\frac{d[\mathrm{N}_2 \mathrm{O}_5]}{dt} = 0.0005[\mathrm{N}_2 \mathrm{O}_5]\). This is a first-order linear differential equation where the rate of decrease of concentration is proportional to the concentration itself.

Step 2: Separate Variables

Rewrite the equation as \(\frac{1}{[\mathrm{N}_2 \mathrm{O}_5]} d[\mathrm{N}_2 \mathrm{O}_5] = -0.0005 \, dt\) to separate variables. This structure allows us to integrate both sides.

Step 3: Integrate Both Sides

Integrate the left side with respect to \([\mathrm{N}_2 \mathrm{O}_5]\) and the right side with respect to \(t\). This gives \(\ln|[\mathrm{N}_2 \mathrm{O}_5]| = -0.0005t + C_1\), where \(C_1\) is the constant of integration.

Step 4: Solve for the Concentration

Solve the equation \(\ln|[\mathrm{N}_2 \mathrm{O}_5]| = -0.0005t + C_1\) for \([\mathrm{N}_2 \mathrm{O}_5]\). Exponentiate both sides to obtain \([\mathrm{N}_2 \mathrm{O}_5] = e^{-0.0005t + C_1} = e^{C_1} e^{-0.0005t}\). Let \(e^{C_1} = C\), the initial concentration, thus \([\mathrm{N}_2 \mathrm{O}_5] = C e^{-0.0005t}\).

Step 5: Determine Time to Original 90% Concentration

To find the time when \([\mathrm{N}_2 \mathrm{O}_5] = 0.9C\), substitute into the equation: \(0.9C = C e^{-0.0005t}\). Simplify to get \(0.9 = e^{-0.0005t}\). Solve for \(t\) by taking the natural logarithm: \(-0.0005t = \ln(0.9)\). Therefore, \(t = \frac{\ln(0.9)}{-0.0005}\).

Step 6: Calculate Numerical Value for Time

Compute the numerical value for \(t\) using a calculator: \(t = \frac{-0.1054}{-0.0005}\), approximating \(t \approx 210.8\) seconds.

Key Concepts

Reaction Rate

The term **reaction rate** refers to the speed at which a chemical reaction occurs. It is quantified by the change in concentration of a reactant or product per unit time. In our exercise, we look at the reaction rate for the decomposition of dinitrogen pentoxide, or \ ext{N}_2 ext{O}_5. In other words, how quickly \ ext{N}_2 ext{O}_5 is consumed or transformed into other substances affects the reaction rate. Here, we're provided with a specific rate formula:

• \(-\frac{d[\text{N}_2 \text{O}_5]}{dt} = 0.0005[\text{N}_2 \text{O}_5]\)

This equation tells us that the reaction rate is directly proportional to the concentration of \ ext{N}_2 ext{O}_5. Meaning, how fast this substance reacts depends linearly on how much of it is present. A higher concentration implies a faster rate and vice versa.

First-order Reactions

First-order reactions are reactions where the rate is directly proportional to only one reactant's concentration. In our context, the decomposition of dinitrogen pentoxide (\( ext{N}_{2} ext{O}_{5}\) ) happens to be first-order. The rate law for such reactions can be described using the formula:

• \(-\frac{d[A]}{dt} = k[A]\)

Here, \([A]\) represents the concentration of the reactant, while \(k\) stands for the reaction rate constant. Our exercise mirrors this with \(k = 0.0005\).Such reactions are important because they exhibit a characteristic exponential decay of concentration over time. This means the time taken for the concentration to halve (termed as half-life) remains constant throughout the reaction. Understanding these dynamics is crucial in predicting how quickly a reaction reaches completion.

Separation of Variables

**Separation of variables** is a method used to solve differential equations. It's particularly useful for simple equations involving derivatives that equate to a product of functions of different variables. It's about separating these variables so each side of the equation contains only one of these variables.In the context of our given problem, we have:

• \(-\frac{d[\text{N}_2 \text{O}_5]}{dt} = 0.0005[\text{N}_2 \text{O}_5]\)

This equation can be rearranged to isolate both variables:

• \(\frac{1}{[\text{N}_2 \text{O}_5]} d[\text{N}_2 \text{O}_5] = -0.0005 dt\)

We now have all terms involving \([\text{N}_2 \text{O}_5]\) on one side of the equation and all terms involving \(t\) on the other.This setup allows us to integrate each side independently, making it simpler to solve for the unknown function involving concentration over time.

Exponential Functions

An important mathematical concept in understanding first-order reactions is the **exponential function**. These functions are often used to describe how quantities grow or decay at a constant rate.In our reaction rate problem, after integrating using separation of variables, we derive the expression:

• \([\text{N}_2 \text{O}_5] = C e^{-0.0005t}\)

The function \(e^{-0.0005t}\) represents exponential decay. Here, \(C\) is the initial concentration, indicating that this function describes how the concentration decreases exponentially over time.Exponential functions are characterized by the presence of \(e\), the base of natural logarithms, which signifies continuous growth or decay. Such functions are crucial in chemistry to model reactions, population biology, and finance to describe how investment grows or shrinks over time. Understanding their behavior helps in predicting how processes that are modeled by these functions unfold over time.