# 4.4: Problems

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##### Problem $$\PageIndex{1}$$

In each case,

• Identify the dependent and independent variables.
• Determine if the differential equation is separable.
• Determine if the differential equation is linear.
• Find the general solution.
• Find the particular solution using the given initial condition.
• Verify that your solution satisfies the differential equation by substitution.
1. $$\frac{dy}{dx}=\frac{y+2}{x-3}, y(0)=1$$
2. $$x'=e^{x+t}, x'(0)=2$$
3. $$\frac{dy}{dx}-\frac{3}{x}y=2x^4, y(1)=1$$
4. $$\frac{df}{dt}=\frac{3t^2}{f}, f(2)=4$$
5. $$h'(t)+2h(t)=4, h(0)=1$$
##### Problem $$\PageIndex{2}$$

Consider the reaction $$A \overset{k}{\rightarrow}B$$. The rate of disappearance of A is proportional to the concentration of A, so:

$-\frac{d[A]}{dt}=k[A] \nonumber$

1) Obtain [A]$$(t)$$ and [B]$$(t)$$.

2) Using the definition of half-life $$(t_{1/2})$$, obtain an expression for $$(t_{1/2})$$ for this mechanism. Your result will be a function of $$k$$.

3) Sketch $$[A](t)$$ and $$[B](t)$$ for the case $$k=0.1 s^{-1}$$, $$[B]_0=0$$ and $$[A]_0=10^{-3}M$$. Remember that you are expected to do this without the help of a calculator.

##### Problem $$\PageIndex{3}$$

Consider the reaction $$2A \overset{k}{\rightarrow}B$$. This mechanism is called a bi-molecular reaction, because the reaction involves the collision of two molecules of reactant. In this case, the rate of disappearance of A is proportional to the square of the concentration of A, so:

$-\frac{1}{2}\frac{d[A]}{dt}=k[A]^2 \nonumber$

Notice that the rate is proportional to the square of the concentration, so this is a second-order reaction.
Assume that the initial concentration of [A] is [A]$$_0$$, and the initial concentration of [B] is zero.

1. Obtain an expression for [A]$$(t)$$.
2. Write down a mass balance (a relationship relating [A](t), [B](t), [A]$$_0$$ and [B]$$_0$$) and obtain [B]$$(t)$$.
3. Using the definition of half-life $$(t_{1/2})$$, obtain an expression for $$(t_{1/2})$$ for this mechanism. Your result will be a function of $$k$$ and [A]$$_0$$.
##### Problem $$\PageIndex{4}$$

Obtain $$[A](t), [B](t),$$ and$$[C](t)$$ for the following mechanism:

$A \overset{k}{\rightarrow}B\overset{k}{\rightarrow}C \nonumber$

Assume $$[A](0)=[A]_0$$, and $$[B](0)=[C](0)=0$$

Note that this problem is identical to the one solved in Section 4.2 but with $$k_1=k_2$$. Be sure you identify the step where the two problems become different.

##### Problem $$\PageIndex{5}$$

Consider the reaction

$A \xrightleftharpoons[k_2]{k_1} B \nonumber$

modeled mathematically by the following ODE

$\frac{d[A]}{dt}=k_2[B]-k_1[A] \nonumber$

The constants, $$k_1$$ and $$k_2$$ represent the kinetic constants in the forward and backward direction respectively, and [A] and [B] represent the molar concentration of A and B. Assume you start with initial concentrations $$[A] (t = 0) = [A]_0$$ and $$[B](t = 0) = [B]_0$$.

Mass conservation requires that $$[A](t) + [B](t) = [A]_0 + [B]_0$$

1. Obtain [A](t) and [B](t) in terms of $$k_1, k_2, [A]_0$$ and $$[B]_0$$.
2. Obtain expressions for the concentrations of A and B in equilibrium: $$[A]_{eq} =[A](t \rightarrow \infty$$) and $$[B]_{eq} =[B](t \rightarrow \infty$$).
3. Prove that the equilibrium constant of the reaction, $$\frac{[B]_{eq}}{[A]_{eq}} = K_{eq}$$, can be expressed as $\frac{[B]_{eq}}{[A]_{eq}} = \frac{k_1}{k_2} \nonumber$
4. Assume that $$k_1=1 min^{-1}$$, $$k_2=\frac{1}{2} min^{-1}$$, $$[A]_0=0$$ and $$[B]_0=0.1 M$$, calculate $$[A]_{eq}$$ and $$[B]_{eq}$$, and sketch $$[A](t)$$ and $$[B](t)$$ to the best of your abilities.

This page titled 4.4: Problems is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Marcia Levitus via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.