4.4: Problems
- Page ID
- 106823
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)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.
- \(\frac{dy}{dx}=\frac{y+2}{x-3}, y(0)=1\)
- \(x'=e^{x+t}, x'(0)=2\)
- \(\frac{dy}{dx}-\frac{3}{x}y=2x^4, y(1)=1\)
- \(\frac{df}{dt}=\frac{3t^2}{f}, f(2)=4\)
- \(h'(t)+2h(t)=4, h(0)=1\)
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.
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.
- Obtain an expression for [A]\((t)\).
- Write down a mass balance (a relationship relating [A](t), [B](t), [A]\(_0\) and [B]\(_0\)) and obtain [B]\((t)\).
- 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]\(_0\).
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.
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 \([A](t) + [B](t) = [A]_0 + [B]_0\)
- Obtain [A](t) and [B](t) in terms of \(k_1, k_2, [A]_0\) and \([B]_0\).
- Obtain expressions for the concentrations of A and B in equilibrium: \([A]_{eq} =[A](t \rightarrow \infty\)) and \([B]_{eq} =[B](t \rightarrow \infty\)).
- Prove that the equilibrium constant of the reaction, \(\frac{[B]_{eq}}{[A]_{eq}} = K_{eq}\), can be expressed as \[\frac{[B]_{eq}}{[A]_{eq}} = \frac{k_1}{k_2} \nonumber \]
- 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.