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Chapter 24: Problem 12

Derive a relation between \(K_{P}\) and \(K_{c}\) for the following: (a) \(\quad \mathrm{CO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{COCl}_{2}(\mathrm{g})\) (b) \(\quad \mathrm{CO}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CH}_{4}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) (c) \(2 \operatorname{BrCl}(g) \rightleftharpoons \operatorname{Br}_{2}(g)+C l_{2}(g)\)

Short Answer

Expert verified
(a) \( K_p = \frac{K_c}{RT} \), (b) \( K_p = \frac{K_c}{(RT)^2} \), (c) \( K_p = K_c \).

Step by step solution

01

Understand the relationship between Kp and Kc

The relationship between the equilibrium constants for gases, \( K_p \) and \( K_c \), is given by the equation: \[ K_p = K_c (RT)^{\Delta n} \] where \( R \) is the ideal gas constant (0.0821 L路atm/mol路K), \( T \) is the temperature in Kelvin, and \( \Delta n \) is the change in the number of moles of gas (moles of products minus moles of reactants).
02

Step 2a: Calculate 螖n for the first reaction

For the reaction \( \mathrm{CO}(\mathrm{g}) + \mathrm{Cl}_2(\mathrm{g}) \rightleftharpoons \mathrm{COCl}_2(\mathrm{g}) \), we calculate \( \Delta n \) as follows:- Moles of products = 1- Moles of reactants = 1 (CO) + 1 (Cl鈧�) = 2- \( \Delta n = 1 - 2 = -1 \)
03

Step 3a: Derive Kp in terms of Kc for the first reaction

Using the formula from Step 1 and \( \Delta n = -1 \):\[ K_p = K_c (RT)^{-1} = \frac{K_c}{RT} \]
04

Step 2b: Calculate 螖n for the second reaction

For the reaction \( \mathrm{CO}(\mathrm{g}) + 3\mathrm{H}_2(\mathrm{g}) \rightleftharpoons \mathrm{CH}_4(\mathrm{g}) + \mathrm{H}_2\mathrm{O}(\mathrm{g}) \), we calculate \( \Delta n \) as follows:- Moles of products = 1 (CH鈧�) + 1 (H鈧侽) = 2- Moles of reactants = 1 (CO) + 3 (H鈧�) = 4- \( \Delta n = 2 - 4 = -2 \)
05

Step 3b: Derive Kp in terms of Kc for the second reaction

Using the formula and \( \Delta n = -2 \):\[ K_p = K_c (RT)^{-2} = \frac{K_c}{(RT)^2} \]
06

Step 2c: Calculate 螖n for the third reaction

For the reaction \( 2\operatorname{BrCl}(g) \rightleftharpoons \operatorname{Br}_2(g) + Cl_2(g) \), calculate \( \Delta n \) as follows:- Moles of products = 1 (Br鈧�) + 1 (Cl鈧�) = 2- Moles of reactants = 2 (BrCl) = 2- \( \Delta n = 2 - 2 = 0 \)
07

Step 3c: Derive Kp in terms of Kc for the third reaction

Since \( \Delta n = 0 \), the relationship between \( K_p \) and \( K_c \) simplifies:\[ K_p = K_c (RT)^0 = K_c \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kp and Kc relationship
Understanding the relationship between the equilibrium constants \( K_p \) and \( K_c \) is essential for dealing with gas reactions in chemistry. \( K_p \) and \( K_c \) represent the equilibrium constants for pressures and concentrations, respectively. The mathematical relationship between them is expressed as: \[ K_p = K_c (RT)^{\Delta n} \]Where:
  • \( R \) is the ideal gas constant.
  • \( T \) is the absolute temperature in Kelvin.
  • \( \Delta n \) is the change in moles of gas, calculated as the moles of gaseous products minus the moles of gaseous reactants.
This relationship shows that \( \Delta n \) plays an important role in determining how \( K_p \) varies from \( K_c \). For reactions where the number of moles of gas does not change, \( \Delta n = 0 \), and thus \( K_p = K_c \). However, when the moles of gases increase or decrease, it affects the relationship, making it vital to compute \( \Delta n \) accurately.
Ideal gas constant
The ideal gas constant, often denoted as \( R \), is a crucial factor in many thermodynamic calculations. It links several physical quantities in the context of the ideal gas law and is a constant used in the equation relating \( K_p \) and \( K_c \):- Its value is typically 0.0821 L路atm/mol路K when pressures are measured in atmospheres and volumes in liters.- It serves as a proportionality factor that relates energy at the molecular level to macroscopic quantities like temperature and pressure.
Because \( R \) is universal for all ideal gases, it allows for the consistent conversion between different types of equilibrium constants across varying conditions. This versatility is what makes \( R \) one of the most powerful tools in a chemist鈥檚 toolkit.
Mole concept
The mole concept is foundational in chemistry, providing a bridge between the atomic and macroscopic worlds. It allows chemists to count atoms and molecules in measurable amounts. When dealing with gas reactions:
  • Understanding how the number of moles changes (i.e., \( \Delta n \)) is key when calculating changes in equilibrium states.
  • The concept is crucial for converting between mass and volume in reactions where gases are involved.
In equilibrium expressions, the mole concept is used to accurately depict how reactants and products interact on a molecular level, allowing for precise computation of equilibrium constants like \( K_c \) and conversion to \( K_p \).
Reaction thermodynamics
Reaction thermodynamics involves understanding how energy changes within a chemical reaction affect equilibrium. The temperature, pressure, and volume are crucial factors that influence the position of equilibrium and the values of equilibrium constants.In the case of \( K_p \) and \( K_c \), reaction thermodynamics come into play as:
  • Temperature affects both \( K_c \) and \( K_p \) through the ideal gas law and the dependence on \( (RT)^{\Delta n} \).
  • The calculations of \( \Delta n \) provide insights into whether products or reactants are favored under specific conditions.
Understanding these dynamics allows chemists to predict the behavior of reactions under different thermodynamic conditions, thus optimizing conditions for desired reactions in both industrial and laboratory settings.

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Most popular questions from this chapter

Consider the reaction described by $$ 3 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{N}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g}) $$ at \(500 \mathrm{~K}\) and a total pressure of one bar. Suppose that we start with three moles of \(\mathrm{H}_{2}(\mathrm{~g})\), one mole of \(\mathrm{N}_{2}(\mathrm{~g})\), and no \(\mathrm{NH}_{3}(\mathrm{~g})\). Show that $$ \begin{aligned} G(\xi)=&(3-3 \xi) G_{\mathrm{H}_{2}}^{\circ}+(1-\xi) G_{\mathrm{N}_{2}}^{\circ}+2 \xi G_{\mathrm{NH}_{3}}^{\circ} \\ &+(3-3 \xi) R T \ln \frac{3-3 \xi}{4-2 \xi}+(1-\xi) R T \ln \frac{1-\xi}{4-2 \xi}+2 \xi R T \ln \frac{2 \xi}{4-2 \xi} \end{aligned} $$ where \(\xi\) is the extent of reaction. Given that \(G_{\mathrm{NH}_{3}}^{\circ}=4.800 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\) at \(500 \mathrm{~K}\) (see Table \(24.4\) ), plot \(G(\xi)\) versus \(\xi\). Differentiate \(G(\xi)\) with respect to \(\xi\) and show that the minimum value of \(G(\xi)\) occurs at \(\xi_{\text {eq }}=0.158\). Also show that $$ \left(\frac{\partial G}{\partial \xi}\right)_{T, P}=\Delta_{\mathrm{r}} G^{\circ}+R T \ln \frac{P_{\mathrm{NH}_{3}}^{2}}{P_{\mathrm{H}_{2}}^{3} P_{\mathrm{N}_{2}}} $$ and that \(K_{\mathrm{p}}=16 \xi_{\mathrm{eq}}^{2}\left(2-\xi_{\mathrm{eq}}\right)^{2} / 27\left(1-\xi_{\mathrm{eq} q}\right)^{4}=0.10\).

Suppose that we have a mixture of the gases \(\mathrm{H}_{2}(\mathrm{g}), \mathrm{CO}_{2}(\mathrm{g}), \mathrm{CO}(\mathrm{g}),\) and \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) at \(1260 \mathrm{K},\) with \(P_{\mathrm{H},}=0.55\) bar, \(P_{\mathrm{CO}_{2}}=0.20 \mathrm{bar}, P_{\mathrm{CO}}=1.25\) bar, and \(P_{\mathrm{H}_{2} \mathrm{O}}=0.10\) bar. Is the reaction described by the equation \\[ \mathrm{H}_{2}(\mathrm{g})+\mathrm{CO}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \quad K_{p}=1.59 \\] at equilibrium under these conditions? If not, in what direction will the reaction proceed to attain equilibrium?

The temperature dependence of the equilibrium constant \(K_{p}\) for the reaction described by \\[ 2 \mathrm{C}_{3} \mathrm{H}_{6}(\mathrm{g}) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{g})+\mathrm{C}_{4} \mathrm{H}_{8}(\mathrm{g}) \\] is given by the equation \\[ \ln K_{p}(T)=-2.395-\frac{2505 \mathrm{K}}{T}+\frac{3.477 \times 10^{6} \mathrm{K}^{2}}{T^{2}} \quad 300 \mathrm{K}

Consider the reaction described by $$ 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons 2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) $$ at \(4000 \mathrm{~K}\) and a total pressure of one bar. Suppose that we start with two moles of \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) and no \(\mathrm{H}_{2}\left(\mathrm{~g}\right.\) ) or \(\mathrm{O}_{2}(\mathrm{~g})\). Show that $$ \begin{gathered} G(\xi)=2(1-\xi) G_{\mathrm{H}_{2} \mathrm{O}}^{\circ}+2 \xi G_{\mathrm{H}_{2}}^{\circ}+\xi G_{\mathrm{O}_{2}}^{\circ}+2(1-\xi) R T \ln \frac{2(1-\xi)}{2+\xi} \\ +2 \xi R T \ln \frac{2 \xi}{2+\xi}+\xi R T \ln \frac{\xi}{2+\xi} \end{gathered} $$ where \(\xi\) is the extent of reaction. Given that \(\Delta_{\mathrm{f}} G^{\circ}\left[\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\right]=-18.334 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\) at \(4000 \mathrm{~K}\), plot \(G(\xi)\) against \(\xi\). Differentiate \(G(\xi)\) with respect to \(\xi\) and show that the minimum value of \(G(\xi)\) occurs at \(\xi_{\mathrm{eq}}=0.553\). Also show that $$ \left(\frac{\partial G}{\partial \xi}\right)_{T, P}=\Delta_{\mathrm{r}} G^{\circ}+R T \ln \frac{P_{\mathrm{H}_{2}}^{2} P_{\mathrm{O}_{2}}}{P_{\mathrm{H}_{2} \mathrm{O}}^{2}} $$ and that \(K_{\mathrm{p}}=\xi_{\mathrm{eq}}^{3} /\left(2+\xi_{\mathrm{eq}}\right)\left(1-\xi_{\mathrm{eq}}\right)^{2}=0.333\).

Consider the dissociation of \(\mathrm{HgO}\left(\mathrm{s}\right.\), red) to \(\mathrm{Hg}(\mathrm{g})\) and \(\mathrm{O}_{2}(\mathrm{~g})\) according to $$ \mathrm{HgO}(\mathrm{s}, \text { red }) \rightleftharpoons \mathrm{Hg}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) $$ If we start with only \(\mathrm{HgO}(\mathrm{s}, \mathrm{red})\), then assuming ideal behavior, show that $$ K_{P}=\frac{2}{3^{3 / 2}} P^{3 / 2} $$ where \(P\) is the total pressure. Given the following "dissociation pressure" of \(\mathrm{HgO}(\mathrm{s}, \mathrm{red})\) at various temperatures, plot \(\ln K_{P}\) versus \(1 / T\). \begin{tabular}{llll} \(t /{ }^{\circ} \mathrm{C}\) & \(\mathrm{P} / \mathrm{atm}\) & \(t /{ }^{\circ} \mathrm{C}\) & \(\mathrm{P} / \mathrm{atm}\) \\ \hline 360 & \(0.1185\) & 430 & \(0.6550\) \\ 370 & \(0.1422\) & 440 & \(0.8450\) \\ 380 & \(0.1858\) & 450 & \(1.067\) \\ 390 & \(0.2370\) & 460 & \(1.339\) \\ 400 & \(0.3040\) & 470 & \(1.674\) \\ 410 & \(0.3990\) & 480 & \(2.081\) \\ 420 & \(0.5095\) & & \end{tabular} An excellent curve fit to the plot of \(\ln K_{P}\) against \(1 / T\) is given by $$ \begin{gathered} \ln K_{P}=-172.94+\frac{4.0222 \times 10^{5} \mathrm{~K}}{T}-\frac{2.9839 \times 10^{8} \mathrm{~K}^{2}}{T^{2}}+\frac{7.0527 \times 10^{10} \mathrm{~K}^{3}}{T^{3}} \\ 630 \mathrm{~K}
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