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Chapter 17: Problem 77

The formation of methanol is important to the processing of new fuels. At \(298 \mathrm{~K}, K_{\mathrm{p}}=2.25 \times 10^{4}\) for the reaction $$ \mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(l) $$ If \(\Delta H_{\mathrm{rxn}}^{\circ}=-128 \mathrm{~kJ} / \mathrm{mol} \mathrm{CH}_{3} \mathrm{OH},\) calculate \(K_{\mathrm{p}}\) at \(0^{\circ} \mathrm{C}\)

Short Answer

Expert verified
Kp at 0掳C is approximately 1.1354x10^17.

Step by step solution

01

- Identify the given data

The data provided are:1. The equilibrium constant at 298 K, denoted as 饾懖饾拺: \(K_{p}(298 \text{ K}) = 2.25 \times 10^{4} \)2. The change in enthalpy of the reaction, denoted as 螖贬掳: \(螖贬掳_{\text{rxn}} = -128 \text{ kJ/mol} \)3. The initial temperature: 298 K4. The final temperature: 0掳C, which converts to 273 K
02

- Utilize the Van鈥檛 Hoff equation

The Van鈥檛 Hoff equation relates the change in the equilibrium constant with temperature: \[\frac{d (\text{ln} K_p)}{dT} = \frac{螖贬掳_{\text{rxn}}}{R T^2} \]
03

- Integrate the Van鈥檛 Hoff equation

To find the equilibrium constant at a temperature different from the initial, integrate the Van't Hoff equation: \[ \frac{d (\text{ln} K_p)}{dT} = \frac{螖贬掳_{\text{rxn}}}{R T^2} \]Integrating both sides with respect to T from T1 to T2 gives: \[ \text{ln} \frac{K_{p2}}{K_{p1}} = -\frac{螖贬掳_{\text{rxn}}}{R} \bigg( \frac{1}{T_2} - \frac{1}{T_1} \bigg) \]
04

- Substitute known values

Substitute the known values:\[ K_{p1} = 2.25 \times 10^4 \ 螖贬掳_{\text{rxn}} = -128 \times 10^3 \text{ J/mol } \ T_1 = 298 \text{ K } \ T_2 = 273 \text{ K } \ R = 8.314 \text{ J/mol路K } \]This yields: \[ \text{ln} \frac{K_{p2}}{2.25 \times 10^4} = -\frac{-128 \times 10^3 \text{ J/mol}}{8.314 \text{ J/mol路K}} \bigg( \frac{1}{273} - \frac{1}{298} \bigg) \]
05

- Calculate the right side

Calculate the expression on the right side of the equation:\[ -\frac{-128 \times 10^3}{8.314} \bigg( \frac{1}{273} - \frac{1}{298} \bigg) \]This value calculates as:\[ \frac{128 \times 10^3}{8.314} \bigg( \frac{1}{273} - \frac{1}{298} \bigg) = 211.33 \ - \frac{1}{273} + \frac{1}{298} = 0.000134 \ 211.33 \times 0.000134 = 28.3 \]
06

- Solve for \(\text{ln} \)

\(\text{ln}\frac{K_{p2}}{2.25 \times 10^4} = 28.3\)To solve for \(\frac{K_{p2}}{2.25 \times 10^4}\), use exponentiation with base \(e\):\(\text{e}^{28.3}\) is a large number; hence \(K_{p2} = (2.25 \times 10^4) \text{e}^{28.3}\).
07

- Final calculation

Calculate \(\text{e}^{28.3}\) and multiply by \(2.25 \times 10^4\).\(\text{e}^{28.3} \approx 5.0463\times10^{12}\)Thus,\(K_{p2} \approx 2.25 \times 10^4 \times 5.0463\times10^{12} = 1.1354\times10^{17}\)

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Key Concepts

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

Thermodynamics
Thermodynamics is a branch of physics that studies the relationships between heat, temperature, and energy. In chemistry, it helps understand how energy changes during chemical reactions.
The Van鈥檛 Hoff equation, which we鈥檒l discuss later, is rooted in thermodynamics. It highlights how equilibrium constants change with temperature.
When a chemical reaction occurs, it either absorbs or releases energy. This energy change is an essential aspect of thermodynamics. Imagine you鈥檙e holding ice. The energy from your warm hand melts the ice, turning it into water. This process involves energy transfer, a key concept in thermodynamics.
  • System: The part of the universe we鈥檙e studying. For our exercise, it鈥檚 the reaction forming methanol.
  • Surroundings: Everything else outside the system.
  • State Functions: Properties like enthalpy (螖贬) and entropy (S) that depend only on the state of the system, not how it got there.
Chemical Equilibrium
Chemical equilibrium occurs when the rate at which reactants transform into products equals the rate at which products revert to reactants. For our methanol formation reaction, equilibrium is reached when the concentrations of CO, H鈧�, and CH鈧僌H levels off.
At equilibrium, the concentration of each reactant and product remains constant. This doesn鈥檛 mean the reactions stop; they simply occur at the same rate in both directions.
An important aspect of equilibrium is the equilibrium constant (K). For gas-phase reactions, we use the equilibrium constant Kp, which relates to the partial pressures of gases involved.
In our exercise, the equilibrium constant Kp at 298 K is given as 2.25 脳 10鲁.
  • If Kp is much greater than 1, products dominate at equilibrium.
  • If Kp is much less than 1, reactants dominate at equilibrium.

Using the Van鈥檛 Hoff equation, we can calculate how Kp changes with temperature. This equation is handy for understanding how equilibrium shifts with varying conditions.
Enthalpy Change
Enthalpy change (\textbf{螖贬}) is the heat absorbed or released during a reaction at constant pressure. It鈥檚 a crucial concept in thermodynamics and our van鈥檛 Hoff equation.
For our methanol formation reaction, the enthalpy change, 螖贬rxn, is -128 kJ/mol. This negative value tells us the reaction is exothermic, meaning it releases heat.
When using the Van鈥檛 Hoff equation, 螖贬 helps determine how the equilibrium constant (Kp) changes with temperature.
Here鈥檚 how 螖贬 relates to the Van鈥檛 Hoff equation: \[ \frac{d (\text{ln} K_{p})}{dT} = \frac{\Delta H^{\circ}_{\text{rxn}}}{R T^{2}} \]
In the integrated form, this becomes: \[ \text{ln} \frac{K_{p2}}{K_{p1}} = -\frac{\Delta H^{\circ}_{\text{rxn}}}{R} \left( \frac{1}{T_{2}} - \frac{1}{T_{1}} \right)\]
By substituting the given and known values:
  • 螖贬rxn as -128 kJ/mol
  • R (gas constant) as 8.314 J/mol路K
  • T鈧� as 298 K and T鈧� as 273 K

We calculate Kp at 0掳C (273 K). Understanding 螖贬 is key to predicting how temperature changes affect equilibrium.

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Most popular questions from this chapter

Determine \(\Delta n_{\text {gas }}\) for each of the following reactions: (a) \(\mathrm{MgCO}_{3}(s) \rightleftharpoons \mathrm{MgO}(s)+\mathrm{CO}_{2}(g)\) (b) \(2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}(l)\) (c) \(\mathrm{HNO}_{3}(l)+\mathrm{ClF}(g) \rightleftharpoons \mathrm{ClONO}_{2}(g)+\mathrm{HF}(g)\)

An engineer examining the oxidation of \(\mathrm{SO}_{2}\) in the manufacture of sulfuric acid determines that \(K_{\mathrm{c}}=1.7 \times 10^{8}\) at \(600 . \mathrm{K}:\) $$ 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g) $$ (a) At equilibrium, \(P_{\mathrm{SO}_{3}}=300 .\) atm and \(P_{\mathrm{O}_{2}}=100 .\) atm. Calculate \(P_{\mathrm{sO}_{2}}\). (b) The engineer places a mixture of \(0.0040 \mathrm{~mol}\) of \(\mathrm{SO}_{2}(g)\) and \(0.0028 \mathrm{~mol}\) of \(\mathrm{O}_{2}(g)\) in a 1.0 -L container and raises the temperature to \(1000 \mathrm{~K}\). At equilibrium, \(0.0020 \mathrm{~mol}\) of \(\mathrm{SO}_{3}(g)\) is present. Calculate \(K_{c}\) and \(P_{\mathrm{SO}_{2}}\) for this reaction at \(1000 . \mathrm{K}\).

For the following equilibrium system, which of the changes will form more \(\mathrm{CaCO}_{3} ?\) $$ \begin{array}{r} \mathrm{CO}_{2}(g)+\mathrm{Ca}(\mathrm{OH})_{2}(s) \rightleftharpoons \mathrm{CaCO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l) \\ \Delta H^{\circ}=-113 \mathrm{~kJ} \end{array} $$ (a) Decrease temperature at constant pressure (no phase change). (b) Increase volume at constant temperature. (c) Increase partial pressure of \(\mathrm{CO}_{2}\). (d) Remove one-half of the initial \(\mathrm{CaCO}_{3}\).

Hydrogen iodide decomposes according to the reaction $$ 2 \mathrm{HI}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) $$ A sealed 1.50-L container initially holds \(0.00623 \mathrm{~mol}\) of \(\mathrm{H}_{2}\), \(0.00414 \mathrm{~mol}\) of \(\mathrm{I}_{2}\), and \(0.0244 \mathrm{~mol}\) of \(\mathrm{HI}\) at \(703 \mathrm{~K}\). When equilibrium is reached, the concentration of \(\mathrm{H}_{2}(g)\) is \(0.00467 \mathrm{M}\). What are the concentrations of \(\mathrm{HI}(g)\) and \(\mathrm{I}_{2}(g) ?\)

Phosgene \(\left(\mathrm{COCl}_{2}\right)\) is a toxic substance that forms readily from carbon monoxide and chlorine at elevated temperatures: $$ \mathrm{CO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{COCl}_{2}(g) $$ If \(0.350 \mathrm{~mol}\) of each reactant is placed in a 0.500 - L flask at \(600 \mathrm{~K}\), what are the concentrations of all three substances at equilibrium \(\left(K_{c}=4.95\right.\) at this temperature)?
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