| [{"problem_text": "The change in molar internal energy when $\\mathrm{CaCO}_3(\\mathrm{~s})$ as calcite converts to another form, aragonite, is $+0.21 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$. Calculate the difference between the molar enthalpy and internal energy changes when the pressure is 1.0 bar given that the densities of the polymorphs are $2.71 \\mathrm{~g} \\mathrm{~cm}^{-3}$ and $2.93 \\mathrm{~g} \\mathrm{~cm}^{-3}$, respectively.", "answer_latex": " -0.28", "answer_number": "-0.28", "unit": " $\\mathrm{~Pa} \\mathrm{~m}^3 \\mathrm{~mol}^{-1}$", "source": "atkins", "problemid": " 2.2", "comment": " ", "solution": "The change in enthalpy when the transition occurs is\r\n$$\r\n\\begin{aligned}\r\n\\Delta H_{\\mathrm{m}} & =H_{\\mathrm{m}}(\\text { aragonite })-H_{\\mathrm{m}}(\\text { calcite }) \\\\\r\n& =\\left\\{U_{\\mathrm{m}}(\\mathrm{a})+p V_{\\mathrm{m}}(\\mathrm{a})\\right\\}-\\left\\{U_{\\mathrm{m}}(\\mathrm{c})+p V_{\\mathrm{m}}(\\mathrm{c})\\right\\} \\\\\r\n& =\\Delta U_{\\mathrm{m}}+p\\left\\{V_{\\mathrm{m}}(\\mathrm{a})-V_{\\mathrm{m}}(\\mathrm{c})\\right\\}\r\n\\end{aligned}\r\n$$\r\nwhere a denotes aragonite and c calcite. It follows by substituting $V_{\\mathrm{m}}=M / \\rho$ that\r\n$$\r\n\\Delta H_{\\mathrm{m}}-\\Delta U_{\\mathrm{m}}=p M\\left(\\frac{1}{\\rho(\\mathrm{a})}-\\frac{1}{\\rho(\\mathrm{c})}\\right)\r\n$$\r\nSubstitution of the data, using $M=100 \\mathrm{~g} \\mathrm{~mol}^{-1}$, gives\r\n$$\r\n\\begin{aligned}\r\n\\Delta H_{\\mathrm{m}}-\\Delta U_{\\mathrm{m}} & =\\left(1.0 \\times 10^5 \\mathrm{~Pa}\\right) \\times\\left(100 \\mathrm{~g} \\mathrm{~mol}^{-1}\\right) \\times\\left(\\frac{1}{2.93 \\mathrm{~g} \\mathrm{~cm}^{-3}}-\\frac{1}{2.71 \\mathrm{~g} \\mathrm{~cm}^{-3}}\\right) \\\\\r\n& =-2.8 \\times 10^5 \\mathrm{~Pa} \\mathrm{~cm}{ }^3 \\mathrm{~mol}^{-1}=-0.28 \\mathrm{~Pa} \\mathrm{~m}^3 \\mathrm{~mol}^{-1}\r\n\\end{aligned}\r\n$$"}, {"problem_text": "Estimate the molar volume of $\\mathrm{CO}_2$ at $500 \\mathrm{~K}$ and 100 atm by treating it as a van der Waals gas.", "answer_latex": " 0.366", "answer_number": "0.366", "unit": " $\\mathrm{dm}^3\\mathrm{~mol}^{-1}$", "source": "atkins", "problemid": " 1.4", "comment": " ", "solution": "According to Table 1.6, $a=3.610 \\mathrm{dm}^6 \\mathrm{~atm} \\mathrm{~mol}^{-2}$ and $b=4.29 \\times 10^{-2} \\mathrm{dm}^3$ $\\mathrm{mol}^{-1}$. Under the stated conditions, $R T / p=0.410 \\mathrm{dm}^3 \\mathrm{~mol}^{-1}$. The coefficients in the equation for $V_{\\mathrm{m}}$ are therefore\r\n$$\r\n\\begin{aligned}\r\n& b+R T / p=0.453 \\mathrm{dm}^3 \\mathrm{~mol}^{-1} \\\\\r\n& a / p=3.61 \\times 10^{-2}\\left(\\mathrm{dm}^3 \\mathrm{~mol}^{-1}\\right)^2 \\\\\r\n& a b / p=1.55 \\times 10^{-3}\\left(\\mathrm{dm}^3 \\mathrm{~mol}^{-1}\\right)^3\r\n\\end{aligned}\r\n$$\r\nTherefore, on writing $x=V_{\\mathrm{m}} /\\left(\\mathrm{dm}^3 \\mathrm{~mol}^{-1}\\right)$, the equation to solve is\r\n$$\r\nx^3-0.453 x^2+\\left(3.61 \\times 10^{-2}\\right) x-\\left(1.55 \\times 10^{-3}\\right)=0\r\n$$\r\nThe acceptable root is $x=0.366$ (Fig. 1.18), which implies that $V_{\\mathrm{m}}=0.366 \\mathrm{dm}^3 \\mathrm{~mol}^{-1}$."}, {"problem_text": "Suppose the concentration of a solute decays exponentially along the length of a container. Calculate the thermodynamic force on the solute at $25^{\\circ} \\mathrm{C}$ given that the concentration falls to half its value in $10 \\mathrm{~cm}$.", "answer_latex": " 17", "answer_number": "17", "unit": " $\\mathrm{kN} \\mathrm{mol}^{-1}$", "source": "atkins", "problemid": " 20.3", "comment": " ", "solution": "The concentration varies with position as\r\n$$\r\nc=c_0 \\mathrm{e}^{-x / \\lambda}\r\n$$\r\nwhere $\\lambda$ is the decay constant. Therefore,\r\n$$\r\n\\frac{\\mathrm{d} c}{\\mathrm{~d} x}=-\\frac{c}{\\lambda}\r\n$$\r\nEquation 20.45 then implies that\r\n$$\r\n\\mathcal{F}=\\frac{R T}{\\lambda}\r\n$$\r\nWe know that the concentration falls to $\\frac{1}{2} c_0$ at $x=10 \\mathrm{~cm}$, so we can find $\\lambda$ from $\\frac{1}{2}=\\mathrm{e}^{-(10 \\mathrm{~cm}) / \\lambda}$. That is $\\lambda=(10 \\mathrm{~cm} / \\ln 2)$. It follows that\r\n$$\r\n\\mathcal{F}=\\left(8.3145 \\mathrm{~J} \\mathrm{~K}^{-1} \\mathrm{~mol}^{-1}\\right) \\times(298 \\mathrm{~K}) \\times \\ln 2 /\\left(1.0 \\times 10^{-1} \\mathrm{~m}\\right)=17 \\mathrm{kN} \\mathrm{mol}^{-1}\r\n$$\r\nwhere we have used $1 \\mathrm{~J}=1 \\mathrm{~N} \\mathrm{~m}$.\r\n"}, {"problem_text": "A container is divided into two equal compartments (Fig. 5.8). One contains $3.0 \\mathrm{~mol} \\mathrm{H}_2(\\mathrm{~g})$ at $25^{\\circ} \\mathrm{C}$; the other contains $1.0 \\mathrm{~mol} \\mathrm{~N}_2(\\mathrm{~g})$ at $25^{\\circ} \\mathrm{C}$. Calculate the Gibbs energy of mixing when the partition is removed. Assume perfect behaviour.\r\n", "answer_latex": " -6.9", "answer_number": "-6.9", "unit": " $\\mathrm{~kJ}$", "source": "atkins", "problemid": " 5.2", "comment": " ", "solution": "Given that the pressure of nitrogen is $p$, the pressure of hydrogen is $3 p$; therefore, the initial Gibbs energy is\r\n$$\r\nG_{\\mathrm{i}}=(3.0 \\mathrm{~mol})\\left\\{\\mu^{\\ominus}\\left(\\mathrm{H}_2\\right)+R T \\ln 3 p\\right\\}+(1.0 \\mathrm{~mol})\\left\\{\\mu^{\\ominus}\\left(\\mathrm{N}_2\\right)+R T \\ln p\\right\\}\r\n$$\r\nWhen the partition is removed and each gas occupies twice the original volume, the partial pressure of nitrogen falls to $\\frac{1}{2} p$ and that of hydrogen falls to $\\frac{3}{2} p$. Therefore, the Gibbs energy changes to\r\n$$\r\nG_{\\mathrm{f}}=(3.0 \\mathrm{~mol})\\left\\{\\mu^{\\ominus}\\left(\\mathrm{H}_2\\right)+R T \\ln \\frac{3}{2} p\\right\\}+(1.0 \\mathrm{~mol})\\left\\{\\mu^{\\ominus}\\left(\\mathrm{N}_2\\right)+R T \\ln \\frac{1}{2} p\\right\\}\r\n$$\r\nThe Gibbs energy of mixing is the difference of these two quantities:\r\n$$\r\n\\begin{aligned}\r\n\\Delta_{\\text {mix }} G & =(3.0 \\mathrm{~mol}) R T \\ln \\left(\\frac{\\frac{3}{2} p}{3 p}\\right)+(1.0 \\mathrm{~mol}) R T \\ln \\left(\\frac{\\frac{1}{2} p}{p}\\right) \\\\\r\n& =-(3.0 \\mathrm{~mol}) R T \\ln 2-(1.0 \\mathrm{~mol}) R T \\ln 2 \\\\\r\n& =-(4.0 \\mathrm{~mol}) R T \\ln 2=-6.9 \\mathrm{~kJ}\r\n\\end{aligned}\r\n$$\r\n"}, {"problem_text": "What is the mean speed, $\\bar{c}$, of $\\mathrm{N}_2$ molecules in air at $25^{\\circ} \\mathrm{C}$ ?", "answer_latex": " 475", "answer_number": "475", "unit": " $\\mathrm{~m} \\mathrm{~s}^{-1}$", "source": "atkins", "problemid": " 20.1", "comment": " ", "solution": "The integral required is\r\n$$\r\n\\begin{aligned}\r\n\\bar{c} & =4 \\pi\\left(\\frac{M}{2 \\pi R T}\\right)^{3 / 2} \\int_0^{\\infty} v^3 \\mathrm{e}^{-M v^2 / 2 R T} \\mathrm{~d} v \\\\\r\n& =4 \\pi\\left(\\frac{M}{2 \\pi R T}\\right)^{3 / 2} \\times \\frac{1}{2}\\left(\\frac{2 R T}{M}\\right)^2=\\left(\\frac{8 R T}{\\pi M}\\right)^{1 / 2}\r\n\\end{aligned}\r\n$$\r\nwhere we have used the standard result from tables of integrals (or software) that\r\n$$\r\n\\int_0^{\\infty} x^3 \\mathrm{e}^{-a x^2} \\mathrm{~d} x=\\frac{1}{2 a^2}\r\n$$\r\nSubstitution of the data then gives\r\n$$\r\n\\bar{c}=\\left(\\frac{8 \\times\\left(8.3141 \\mathrm{~J} \\mathrm{~K}^{-1} \\mathrm{~mol}^{-1}\\right) \\times(298 \\mathrm{~K})}{\\pi \\times\\left(28.02 \\times 10^{-3} \\mathrm{~kg} \\mathrm{~mol}^{-1}\\right)}\\right)^{1 / 2}=475 \\mathrm{~m} \\mathrm{~s}^{-1}\r\n$$\r\nwhere we have used $1 \\mathrm{~J}=1 \\mathrm{~kg} \\mathrm{~m}^2 \\mathrm{~s}^{-2}$.\r\n"}, {"problem_text": "Caesium (m.p. $29^{\\circ} \\mathrm{C}$, b.p. $686^{\\circ} \\mathrm{C}$ ) was introduced into a container and heated to $500^{\\circ} \\mathrm{C}$. When a hole of diameter $0.50 \\mathrm{~mm}$ was opened in the container for $100 \\mathrm{~s}$, a mass loss of $385 \\mathrm{mg}$ was measured. Calculate the vapour pressure of liquid caesium at $500 \\mathrm{~K}$.", "answer_latex": " 8.7", "answer_number": "8.7", "unit": " $\\mathrm{kPa}$", "source": "atkins", "problemid": " 20.2", "comment": " ", "solution": " The mass loss $\\Delta m$ in an interval $\\Delta t$ is related to the collision flux by\r\n$$\r\n\\Delta m=Z_{\\mathrm{W}} A_0 m \\Delta t\r\n$$\r\nwhere $A_0$ is the area of the hole and $m$ is the mass of one atom. It follows that\r\n$$\r\nZ_{\\mathrm{W}}=\\frac{\\Delta m}{A_0 m \\Delta t}\r\n$$\r\nBecause $Z_{\\mathrm{W}}$ is related to the pressure by eqn 20.14 , we can write\r\n$$\r\np=\\left(\\frac{2 \\pi R T}{M}\\right)^{1 / 2} \\frac{\\Delta m}{A_0 \\Delta t}\r\n$$\r\nBecause $M=132.9 \\mathrm{~g} \\mathrm{~mol}^{-1}$, substitution of the data gives $p=8.7 \\mathrm{kPa}$ (using $1 \\mathrm{~Pa}=$ $\\left.1 \\mathrm{~N} \\mathrm{~m}^{-2}=1 \\mathrm{~J} \\mathrm{~m}^{-1}\\right)$."}, {"problem_text": "To get an idea of the distance dependence of the tunnelling current in STM, suppose that the wavefunction of the electron in the gap between sample and needle is given by $\\psi=B \\mathrm{e}^{-\\kappa x}$, where $\\kappa=\\left\\{2 m_{\\mathrm{e}}(V-E) / \\hbar^2\\right\\}^{1 / 2}$; take $V-E=2.0 \\mathrm{eV}$. By what factor would the current drop if the needle is moved from $L_1=0.50 \\mathrm{~nm}$ to $L_2=0.60 \\mathrm{~nm}$ from the surface?", "answer_latex": " 0.23", "answer_number": "0.23", "unit": " ", "source": "atkins", "problemid": " 8.2", "comment": " ", "solution": " When $L=L_1=0.50 \\mathrm{~nm}$ and $V-E=2.0 \\mathrm{eV}=3.20 \\times 10^{-19} \\mathrm{~J}$ the value of $\\kappa L$ is\r\n$$\r\n\\begin{aligned}\r\n\\kappa L_1 & =\\left\\{\\frac{2 m_{\\mathrm{e}}(V-E)}{\\hbar^2}\\right\\}^{1 / 2} L_1 \\\\\r\n& =\\left\\{\\frac{2 \\times\\left(9.109 \\times 10^{-31} \\mathrm{~kg}\\right) \\times\\left(3.20 \\times 10^{-19} \\mathrm{~J}\\right)}{\\left(1.054 \\times 10^{-34} \\mathrm{~J} \\mathrm{~s}\\right)^2}\\right\\}^{1 / 2} \\times\\left(5.0 \\times 10^{-10} \\mathrm{~m}\\right) \\\\\r\n& =\\left(7.25 \\times 10^9 \\mathrm{~m}^{-1}\\right) \\times\\left(5.0 \\times 10^{-10} \\mathrm{~m}\\right)=3.6\r\n\\end{aligned}\r\n$$\r\nBecause $\\kappa L_1>1$, we use eqn $8.19 \\mathrm{~b}$ to calculate the transmission probabilities at the two distances. It follows that\r\n$$\r\n\\text { current at } \\begin{aligned}\r\n\\frac{L_2}{\\text { current at } L_1} & =\\frac{T\\left(L_2\\right)}{T\\left(L_1\\right)}=\\frac{16 \\varepsilon(1-\\varepsilon) \\mathrm{e}^{-2 \\kappa L_2}}{16 \\varepsilon(1-\\varepsilon) \\mathrm{e}^{-2 \\kappa L_1}}=\\mathrm{e}^{-2 \\kappa\\left(L_2-L_1\\right)} \\\\\r\n& =\\mathrm{e}^{-2 \\times\\left(7.25 \\times 10^{-9} \\mathrm{~m}^{-1}\\right) \\times\\left(1.0 \\times 10^{-10} \\mathrm{~m}\\right)}=0.23\r\n\\end{aligned}\r\n$$\r\n"}, {"problem_text": "Calculate the typical wavelength of neutrons that have reached thermal equilibrium with their surroundings at $373 \\mathrm{~K}$.\r\n", "answer_latex": " 226", "answer_number": "226", "unit": " $\\mathrm{pm}$", "source": "atkins", "problemid": " 19.4", "comment": " ", "solution": "From the equipartition principle, we know that the mean translational kinetic energy of a neutron at a temperature $T$ travelling in the $x$-direction is $E_{\\mathrm{k}}=\\frac{1}{2} k T$. The kinetic energy is also equal to $p^2 / 2 m$, where $p$ is the momentum of the neutron and $m$ is its mass. Hence, $p=(m k T)^{1 / 2}$. It follows from the de Broglie relation $\\lambda=h / p$ that the neutron's wavelength is\r\n$$\r\n\\lambda=\\frac{h}{(m k T)^{1 / 2}}\r\n$$\r\nTherefore, at $373 \\mathrm{~K}$,\r\n$$\r\n\\begin{aligned}\r\n\\lambda & =\\frac{6.626 \\times 10^{-34} \\mathrm{~J} \\mathrm{~s}}{\\left\\{\\left(1.675 \\times 10^{-27} \\mathrm{~kg}\\right) \\times\\left(1.381 \\times 10^{-23} \\mathrm{~J} \\mathrm{~K}^{-1}\\right) \\times(373 \\mathrm{~K})\\right\\}^{1 / 2}} \\\\\r\n& =\\frac{6.626 \\times 10^{-34} \\mathrm{~J} \\mathrm{~s}}{\\left(1.675 \\times 1.381 \\times 373 \\times 10^{-50}\\right)^{1 / 2}\\left(\\mathrm{~kg}^2 \\mathrm{~m}^2 \\mathrm{~s}^{-2}\\right)^{1 / 2}} \\\\\r\n& =2.26 \\times 10^{-10} \\mathrm{~m}=226 \\mathrm{pm}\r\n\\end{aligned}"}, {"problem_text": "Calculate the separation of the $\\{123\\}$ planes of an orthorhombic unit cell with $a=0.82 \\mathrm{~nm}, b=0.94 \\mathrm{~nm}$, and $c=0.75 \\mathrm{~nm}$.", "answer_latex": " 0.21", "answer_number": "0.21", "unit": " $\\mathrm{~nm}$", "source": "atkins", "problemid": " 19.1", "comment": " ", "solution": " For the first part, simply substitute the information into eqn 19.3. For the second part, instead of repeating the calculation, note that, if all three Miller indices are multiplied by $n$, then their separation is reduced by that factor (Fig. 19.12):\r\n$$\r\n\\frac{1}{d_{n h, n k, n l}^2}=\\frac{(n h)^2}{a^2}+\\frac{(n k)^2}{b^2}+\\frac{(n l)^2}{c^2}=n^2\\left(\\frac{h^2}{a^2}+\\frac{k^2}{b^2}+\\frac{l^2}{c^2}\\right)=\\frac{n^2}{d_{h k l}^2}\r\n$$\r\nwhich implies that\r\n$$\r\nd_{n h, n k, n l}=\\frac{d_{h k l}}{n}\r\n$$\r\nAnswer Substituting the indices into eqn 19.3 gives\r\n$$\r\n\\frac{1}{d_{123}^2}=\\frac{1^2}{(0.82 \\mathrm{~nm})^2}+\\frac{2^2}{(0.94 \\mathrm{~nm})^2}+\\frac{3^2}{(0.75 \\mathrm{~nm})^2}=0.22 \\mathrm{~nm}^{-2}\r\n$$\r\nHence, $d_{123}=0.21 \\mathrm{~nm}$.\r\n"}, {"problem_text": "Calculate the moment of inertia of an $\\mathrm{H}_2 \\mathrm{O}$ molecule around the axis defined by the bisector of the $\\mathrm{HOH}$ angle (3). The $\\mathrm{HOH}$ bond angle is $104.5^{\\circ}$ and the bond length is $95.7 \\mathrm{pm}$.", "answer_latex": " 1.91", "answer_number": "1.91", "unit": " $10^{-47} \\mathrm{~kg} \\mathrm{~m}^2$", "source": "atkins", "problemid": "12.1", "comment": " ", "solution": "From eqn 12.2,\r\n$$\r\nI=\\sum_i m_i x_i^2=m_{\\mathrm{H}} x_{\\mathrm{H}}^2+0+m_{\\mathrm{H}} x_{\\mathrm{H}}^2=2 m_{\\mathrm{H}} x_{\\mathrm{H}}^2\r\n$$\r\nIf the bond angle of the molecule is denoted $2 \\phi$ and the bond length is $R$, trigonometry gives $x_{\\mathrm{H}}=R \\sin \\phi$. It follows that\r\n$$\r\nI=2 m_{\\mathrm{H}} R^2 \\sin ^2 \\phi\r\n$$\r\nSubstitution of the data gives\r\n$$\r\n\\begin{aligned}\r\nI & =2 \\times\\left(1.67 \\times 10^{-27} \\mathrm{~kg}\\right) \\times\\left(9.57 \\times 10^{-11} \\mathrm{~m}\\right)^2 \\times \\sin ^2\\left(\\frac{1}{2} \\times 104.5^{\\circ}\\right) \\\\\r\n& =1.91 \\times 10^{-47} \\mathrm{~kg} \\mathrm{~m}^2\r\n\\end{aligned}\r\n$$\r\nNote that the mass of the $\\mathrm{O}$ atom makes no contribution to the moment of inertia for this mode of rotation as the atom is immobile while the $\\mathrm{H}$ atoms circulate around it.\r\n"}, {"problem_text": "The data below show the temperature variation of the equilibrium constant of the reaction $\\mathrm{Ag}_2 \\mathrm{CO}_3(\\mathrm{~s}) \\rightleftharpoons \\mathrm{Ag}_2 \\mathrm{O}(\\mathrm{s})+\\mathrm{CO}_2(\\mathrm{~g})$. Calculate the standard reaction enthalpy of the decomposition.\r\n$\\begin{array}{lllll}T / \\mathrm{K} & 350 & 400 & 450 & 500 \\\\ K & 3.98 \\times 10^{-4} & 1.41 \\times 10^{-2} & 1.86 \\times 10^{-1} & 1.48\\end{array}$", "answer_latex": " +80", "answer_number": "+80", "unit": " $\\mathrm{~kJ} \\mathrm{~mol}^{-1}$", "source": "atkins", "problemid": " 6.3", "comment": " ", "solution": "We draw up the following table:\r\n$\\begin{array}{lllll}T / \\mathrm{K} & 350 & 400 & 450 & 500 \\\\ \\left(10^3 \\mathrm{~K}\\right) / T & 2.86 & 2.50 & 2.22 & 2.00 \\\\ -\\ln K & 7.83 & 4.26 & 1.68 & -0.39\\end{array}$\r\nThese points are plotted in Fig. 6.9. The slope of the graph is $+9.6 \\times 10^3$, so\r\n$$\r\n\\Delta_{\\mathrm{r}} H^{\\ominus}=\\left(+9.6 \\times 10^3 \\mathrm{~K}\\right) \\times R=+80 \\mathrm{~kJ} \\mathrm{~mol}^{-1}\r\n$$"}, {"problem_text": "The osmotic pressures of solutions of poly(vinyl chloride), PVC, in cyclohexanone at $298 \\mathrm{~K}$ are given below. The pressures are expressed in terms of the heights of solution (of mass density $\\rho=0.980 \\mathrm{~g} \\mathrm{~cm}^{-3}$ ) in balance with the osmotic pressure. Determine the molar mass of the polymer.\r\n$\\begin{array}{llllll}c /\\left(\\mathrm{g} \\mathrm{dm}^{-3}\\right) & 1.00 & 2.00 & 4.00 & 7.00 & 9.00 \\\\ h / \\mathrm{cm} & 0.28 & 0.71 & 2.01 & 5.10 & 8.00\\end{array}$\r\n", "answer_latex": " 1.2", "answer_number": "1.2", "unit": " $10^2 \\mathrm{~kg} \\mathrm{~mol}^{-1}$", "source": "atkins", "problemid": " 5.4", "comment": " ", "solution": "The data give the following values for the quantities to plot:\r\n$$\r\n\\begin{array}{llllll}\r\nc /\\left(\\mathrm{g} \\mathrm{dm}^{-3}\\right) & 1.00 & 2.00 & 4.00 & 7.00 & 9.00 \\\\\r\n(h / c) /\\left(\\mathrm{cm} \\mathrm{g}^{-1} \\mathrm{dm}^3\\right) & 0.28 & 0.36 & 0.503 & 0.729 & 0.889\r\n\\end{array}\r\n$$\r\nThe points are plotted in Fig. 5.28. The intercept is at 0.21 . Therefore,\r\n$$\r\n\\begin{aligned}\r\nM & =\\frac{R T}{\\rho g} \\times \\frac{1}{0.21 \\mathrm{~cm} \\mathrm{~g}^{-1} \\mathrm{dm}^3} \\\\\r\n& =\\frac{\\left(8.3145 \\mathrm{~J} \\mathrm{~K}^{-1} \\mathrm{~mol}^{-1}\\right) \\times(298 \\mathrm{~K})}{\\left(980 \\mathrm{~kg} \\mathrm{~m}^{-1}\\right) \\times\\left(9.81 \\mathrm{~m} \\mathrm{~s}^{-2}\\right)} \\times \\frac{1}{2.1 \\times 10^{-3} \\mathrm{~m}^4 \\mathrm{~kg}^{-1}} \\\\\r\n& =1.2 \\times 10^2 \\mathrm{~kg} \\mathrm{~mol}^{-1}\r\n\\end{aligned}\r\n$$"}, {"problem_text": "Estimate the wavelength of electrons that have been accelerated from rest through a potential difference of $40 \\mathrm{kV}$.", "answer_latex": " 6.1", "answer_number": "6.1", "unit": " $10^{-12} \\mathrm{~m}$", "source": "atkins", "problemid": " 7.2", "comment": " ", "solution": " The expression $p^2 / 2 m_{\\mathrm{e}}=e \\Delta \\phi$ solves to $p=\\left(2 m_{\\mathrm{e}} e \\Delta \\phi\\right)^{1 / 2}$; then, from the de Broglie relation $\\lambda=h / p$,\r\n$$\r\n\\lambda=\\frac{h}{\\left(2 m_{\\mathrm{e}} e \\Delta \\phi\\right)^{1 / 2}}\r\n$$\r\nSubstitution of the data and the fundamental constants (from inside the front cover) gives\r\n$$\r\n\\begin{aligned}\r\n\\lambda & =\\frac{6.626 \\times 10^{-34} \\mathrm{~J} \\mathrm{~s}}{\\left\\{2 \\times\\left(9.109 \\times 10^{-31} \\mathrm{~kg}\\right) \\times\\left(1.602 \\times 10^{-19} \\mathrm{C}\\right) \\times\\left(4.0 \\times 10^4 \\mathrm{~V}\\right)\\right\\}^{1 / 2}} \\\\\r\n& =6.1 \\times 10^{-12} \\mathrm{~m}\r\n\\end{aligned}\r\n$$\r\n"}, {"problem_text": "The standard enthalpy of formation of $\\mathrm{H}_2 \\mathrm{O}(\\mathrm{g})$ at $298 \\mathrm{~K}$ is $-241.82 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$. Estimate its value at $100^{\\circ} \\mathrm{C}$ given the following values of the molar heat capacities at constant pressure: $\\mathrm{H}_2 \\mathrm{O}(\\mathrm{g}): 33.58 \\mathrm{~J} \\mathrm{~K}^{-1} \\mathrm{~mol}^{-1} ; \\mathrm{H}_2(\\mathrm{~g}): 28.82 \\mathrm{~J} \\mathrm{~K}^{-1} \\mathrm{~mol}^{-1} ; \\mathrm{O}_2(\\mathrm{~g})$ : $29.36 \\mathrm{~J} \\mathrm{~K}^{-1} \\mathrm{~mol}^{-1}$. Assume that the heat capacities are independent of temperature.", "answer_latex": "-242.6 ", "answer_number": "-242.6", "unit": " $\\mathrm{~kJ} \\mathrm{~mol}^{-1}$", "source": "atkins", "problemid": " 2.6", "comment": " ", "solution": "The reaction is $\\mathrm{H}_2(\\mathrm{~g})+\\frac{1}{2} \\mathrm{O}_2(\\mathrm{~g}) \\rightarrow \\mathrm{H}_2 \\mathrm{O}(\\mathrm{g})$, so\r\n$$\r\n\\Delta_{\\mathrm{r}} C_p^{\\ominus}=C_{p, \\mathrm{~m}}^{\\ominus}\\left(\\mathrm{H}_2 \\mathrm{O}, \\mathrm{g}\\right)-\\left\\{C_{p, \\mathrm{~m}}^{\\ominus}\\left(\\mathrm{H}_2, \\mathrm{~g}\\right)+\\frac{1}{2} C_{p, \\mathrm{~m}}^{\\ominus}\\left(\\mathrm{O}_2, \\mathrm{~g}\\right)\\right\\}=-9.92 \\mathrm{~J} \\mathrm{~K}^{-1} \\mathrm{~mol}^{-1}\r\n$$\r\nIt then follows that\r\n$$\r\n\\Delta_{\\mathrm{r}} H^{\\ominus}(373 \\mathrm{~K})=-241.82 \\mathrm{~kJ} \\mathrm{~mol}^{-1}+(75 \\mathrm{~K}) \\times\\left(-9.92 \\mathrm{~J} \\mathrm{~K}^{-1} \\mathrm{~mol}^{-1}\\right)=-242.6 \\mathrm{~kJ} \\mathrm{~mol}^{-1}\r\n$$\r\n"}, {"problem_text": "The standard potential of the cell $\\mathrm{Pt}(\\mathrm{s})\\left|\\mathrm{H}_2(\\mathrm{~g})\\right| \\mathrm{HBr}(\\mathrm{aq})|\\operatorname{AgBr}(\\mathrm{s})| \\mathrm{Ag}(\\mathrm{s})$ was measured over a range of temperatures, and the data were found to fit the following polynomial:\r\n$$\r\nE_{\\text {cell }}^{\\bullet} / \\mathrm{V}=0.07131-4.99 \\times 10^{-4}(T / \\mathrm{K}-298)-3.45 \\times 10^{-6}(\\mathrm{~T} / \\mathrm{K}-298)^2\r\n$$\r\nThe cell reaction is $\\operatorname{AgBr}(\\mathrm{s})+\\frac{1}{2} \\mathrm{H}_2(\\mathrm{~g}) \\rightarrow \\mathrm{Ag}(\\mathrm{s})+\\mathrm{HBr}(\\mathrm{aq})$. Evaluate the standard reaction Gibbs energy, enthalpy, and entropy at $298 \\mathrm{~K}$.\r\n", "answer_latex": " -21.2", "answer_number": "-21.2", "unit": " $\\mathrm{~kJ} \\mathrm{~mol}^{-1}$", "source": "atkins", "problemid": " 6.5", "comment": " ", "solution": "At $T=298 \\mathrm{~K}, E_{\\text {cell }}^{\\ominus}=+0.07131 \\mathrm{~V}$, so\r\n$$\r\n\\begin{aligned}\r\n\\Delta_{\\mathrm{r}} G^{\\ominus} & =-v F E_{\\text {cell }}^{\\ominus}=-(1) \\times\\left(9.6485 \\times 10^4 \\mathrm{Cmol}^{-1}\\right) \\times(+0.07131 \\mathrm{~V}) \\\\\r\n& =-6.880 \\times 10^3 \\mathrm{~V} \\mathrm{Cmol}^{-1}=-6.880 \\mathrm{~kJ} \\mathrm{~mol}^{-1}\r\n\\end{aligned}\r\n$$\r\nThe temperature coefficient of the cell potential is\r\n$$\r\n\\frac{\\mathrm{d} E_{\\text {cell }}^{\\ominus}}{\\mathrm{d} T}=-4.99 \\times 10^{-4} \\mathrm{~V} \\mathrm{~K}^{-1}-2\\left(3.45 \\times 10^{-6}\\right)(T / \\mathrm{K}-298) \\mathrm{V} \\mathrm{K}^{-1}\r\n$$\r\nAt $T=298 \\mathrm{~K}$ this expression evaluates to\r\n$$\r\n\\frac{\\mathrm{d} E_{\\text {cell }}^{\\ominus}}{\\mathrm{d} T}=-4.99 \\times 10^{-4} \\mathrm{~V} \\mathrm{~K}^{-1}\r\n$$\r\nSo, from eqn 6.36 , the reaction entropy is\r\n$$\r\n\\begin{aligned}\r\n\\Delta_{\\mathrm{r}} S^{\\ominus} & =1 \\times\\left(9.6485 \\times 10^4 \\mathrm{Cmol}^{-1}\\right) \\times\\left(-4.99 \\times 10^{-4} \\mathrm{~V} \\mathrm{~K}^{-1}\\right) \\\\\r\n& =-48.1 \\mathrm{~J} \\mathrm{~K}^{-1} \\mathrm{~mol}^{-1}\r\n\\end{aligned}\r\n$$\r\nThe negative value stems in part from the elimination of gas in the cell reaction. It then follows that\r\n$$\r\n\\begin{aligned}\r\n\\Delta_{\\mathrm{r}} H^{\\ominus} & =\\Delta_{\\mathrm{r}} G^{\\ominus}+T \\Delta_{\\mathrm{r}} S^{\\ominus}=-6.880 \\mathrm{~kJ} \\mathrm{~mol}^{-1}+(298 \\mathrm{~K}) \\times\\left(-0.0482 \\mathrm{~kJ} \\mathrm{~K}^{-1} \\mathrm{~mol}^{-1}\\right) \\\\\r\n& =-21.2 \\mathrm{~kJ} \\mathrm{~mol}^{-1}\r\n\\end{aligned}\r\n$$\r\n"}, {"problem_text": "In an industrial process, nitrogen is heated to $500 \\mathrm{~K}$ in a vessel of constant volume. If it enters the vessel at $100 \\mathrm{~atm}$ and $300 \\mathrm{~K}$, what pressure would it exert at the working temperature if it behaved as a perfect gas?", "answer_latex": " 167", "answer_number": "167", "unit": " $\\mathrm{atm}$", "source": "atkins", "problemid": " 1.2", "comment": " ", "solution": "Cancellation of the volumes (because $V_1=V_2$ ) and amounts (because $\\left.n_1=n_2\\right)$ on each side of the combined gas law results in\r\n$$\r\n\\frac{p_1}{T_1}=\\frac{p_2}{T_2}\r\n$$\r\nwhich can be rearranged into\r\n$$\r\np_2=\\frac{T_2}{T_1} \\times p_1\r\n$$\r\nSubstitution of the data then gives\r\n$$\r\np_2=\\frac{500 \\mathrm{~K}}{300 \\mathrm{~K}} \\times(100 \\mathrm{~atm})=167 \\mathrm{~atm}\r\n$$\r\n"}] |