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Chapter 17: Problem 76

The molecule \(D_{2}\) (where \(D\), deuterium, is \({ }^{2} \mathrm{H}\) ) undergoes a reaction with ordinary \(\mathrm{H}_{2}\) that leads to isotopic equilibrium: $$ \mathrm{D}_{2}(g)+\mathrm{H}_{2}(g) \Longrightarrow 2 \mathrm{DH}(g) \quad K_{\mathrm{p}}=1.80 \text { at } 298 \mathrm{~K} $$ If \(\Delta H_{\mathrm{rxn}}^{\circ}\) is \(0.32 \mathrm{~kJ} / \mathrm{mol} \mathrm{DH},\) calculate \(K_{\mathrm{p}}\) at \(500 . \mathrm{K}\)

Short Answer

Expert verified
K_p at 500 K is 1.90.

Step by step solution

01

Understand the Problem

The reaction being considered is \( \text{D}_2(g) + \text{H}_2(g) \rightarrow 2 \text{DH}(g) \). We are given the equilibrium constant \( K_p \) at 298 K (1.80) and need to find \( K_p \) at 500 K. We also have the standard enthalpy change \( \text{螖H}_{\text{rxn}}^{\text{掳}} \) for the reaction, which is \( 0.32 \text{ kJ/mol} \text{ DH}\).
02

Use the Van't Hoff Equation

The Van't Hoff equation relates the equilibrium constants at two different temperatures: \[ \frac{K_{p2}}{K_{p1}} = \text{exp} \biggl( \frac{螖H_{rxn}^\text{掳}}{R} \biggl( \frac{1}{T_1} - \frac{1}{T_2} \biggr) \biggr) \]. Here, \( K_{p1} \) is the equilibrium constant at the initial temperature \( T_1 \) (298 K), \( K_{p2} \) is the equilibrium constant at the final temperature \( T_2 \) (500 K), \( 螖H_{rxn}^{\text{掳}} \) is the standard enthalpy change (0.32 kJ/mol), and \( R \) is the gas constant (8.314 J/mol路K).
03

Convert 螖H_{rxn}^{\text{掳}} to Joules

Convert the enthalpy change from kJ to J: \( \text{螖H}_{\text{rxn}}^{\text{掳}} = 0.32 \text{ kJ/mol} = 320 \text{ J/mol} \).
04

Substitute Values

Substitute the known values into the Van't Hoff equation: \[ \frac{K_{p2}}{1.80} = \text{exp} \biggl( \frac{320 \text{ J/mol}}{8.314 \text{ J/mol路K}} \biggl( \frac{1}{298 \text{ K}} - \frac{1}{500 \text{ K}} \biggr) \biggr) \]
05

Perform Calculations

Calculate the exponent part first: \[ \frac{320}{8.314} \biggl( \frac{1}{298} - \frac{1}{500} \biggr) = 38.48 \biggl( 0.003356 - 0.002 \biggr) = 38.48 \times 0.001356 = 0.05215 \]. Then compute the exponential term: \( e^{0.05215} = 1.0535 \). Finally solve for \( K_{p2} \): \[ 1.80 \times 1.0535 = 1.896 \]

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Key Concepts

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

Equilibrium Constant
The equilibrium constant, often denoted as either \(K_c\) or \(K_p\), represents the ratio of the concentration of products to reactants at equilibrium, each raised to the power of their respective coefficients in the balanced chemical equation. For reactions in the gas phase, \(K_p\) is frequently used where the concentrations are represented as partial pressures.
The equilibrium constant indicates the extent of a reaction at a given temperature.
  • If \(K > 1\), the reaction favors the formation of products.
  • If \(K < 1\), the reaction favors the formation of reactants.
In the provided exercise, we calculated \(K_p\) at different temperatures using the Van't Hoff equation.
Enthalpy Change
Enthalpy change, symbolized as \( \text{螖H} \), measures the heat change at constant pressure. Standard enthalpy change of reaction, denoted as \( \text{螖H}_{\text{rxn}}^{\text{掳}} \), is critical in determining how heat energy is absorbed or released.
  • A negative \( \text{螖H} \text{ (exothermic)} \) indicates heat release.
  • A positive \( \text{螖H} \text{ (endothermic)} \) indicates heat absorption.
In our problem, the enthalpy change for the reaction \( \text{D}_2(g) + \text{H}_2(g) \rightarrow 2 \text{DH}(g) \) was given as \(0.32 \text{ kJ/mol} \text{ DH} \). We converted this value to Joules (320 J/mol) to use it in the Van't Hoff equation.
Temperature Dependence of Equilibrium
The Van't Hoff equation shows how equilibrium constants change with temperature, making it a powerful tool in predicting reaction behavior. The Van't Hoff equation is:

\[ \frac{K_{p2}}{K_{p1}}=\text{exp} \biggl( \frac{螖H_{\text{rxn}}^\text{掳}}{R} \biggl( \frac{1}{T_{1}} - \frac{1}{T_{2}} \biggr) \biggr) \biggr) \]
Here, \( T_1\) and \( T_2\) are the initial and final temperatures, \( K_{p1} \) and \( K_{p2} \) are the corresponding equilibrium constants, and \( R \) is the gas constant (8.314 J/mol路K).

By understanding this equation, we determined how the equilibrium constant \( K_p \) changes from 298 K to 500 K based on the given enthalpy change. We substituted the values to find that \( K_p \) at 500 K is approximately 1.896.

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Most popular questions from this chapter

The water-gas shift reaction plays a central role in the chemical methods for obtaining cleaner fuels from coal: $$ \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g) $$ At a given temperature, \(K_{\mathrm{p}}=2.7 .\) If \(0.13 \mathrm{~mol}\) of \(\mathrm{CO}, 0.56 \mathrm{~mol}\) of \(\mathrm{H}_{2} \mathrm{O}, 0.62 \mathrm{~mol}\) of \(\mathrm{CO}_{2},\) and \(0.43 \mathrm{~mol}\) of \(\mathrm{H}_{2}\) are put in a \(2.0-\mathrm{L}\) flask, in which direction does the reaction proceed?

Predict the effect of decreasing the container volume on the amounts of each reactant and product in the following reactions: (a) \(\mathrm{C}_{3} \mathrm{H}_{8}(g)+5 \mathrm{O}_{2}(g) \rightleftharpoons 3 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(l)\) (b) \(4 \mathrm{NH}_{3}(g)+3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{~N}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)\)

A key step in the extraction of iron from its ore is \(\mathrm{FeO}(s)+\mathrm{CO}(g) \rightleftharpoons \mathrm{Fe}(s)+\mathrm{CO}_{2}(g) \quad K_{\mathrm{p}}=0.403\) at \(1000^{\circ} \mathrm{C}\) This step occurs in the \(700^{\circ} \mathrm{C}\) to \(1200^{\circ} \mathrm{C}\) zone within a blast furnace. What are the equilibrium partial pressures of \(\mathrm{CO}(g)\) and \(\mathrm{CO}_{2}(g)\) when \(1.00 \mathrm{~atm}\) of \(\mathrm{CO}(g)\) and excess \(\mathrm{FeO}(s)\) react in a sealed container at \(1000^{\circ} \mathrm{C} ?\)

Isolation of Group \(8 \mathrm{~B}(10)\) elements, used as industrial catalysts, involves a series of steps. For nickel, the sulfide ore is roasted in air: \(\mathrm{Ni}_{3} \mathrm{~S}_{2}(s)+\mathrm{O}_{2}(g) \rightleftharpoons \mathrm{NiO}(s)+\mathrm{SO}_{2}(g) .\) The metal oxide is reduced by the \(\mathrm{H}_{2}\) in water gas \(\left(\mathrm{CO}+\mathrm{H}_{2}\right)\) to impure \(\mathrm{Ni}: \mathrm{NiO}(s)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{Ni}(s)+\mathrm{H}_{2} \mathrm{O}(g) .\) The \(\mathrm{CO}\) in water gas then reacts with the metal in the Mond process to form gaseous nickel carbonyl, \(\mathrm{Ni}(s)+\mathrm{CO}(g) \rightleftharpoons \mathrm{Ni}(\mathrm{CO})_{4}(g),\) which is sub- sequently decomposed to the metal. (a) Balance each of the three steps, and obtain an overall balanced equation for the conversion of \(\mathrm{Ni}_{3} \mathrm{~S}_{2}\) to \(\mathrm{Ni}(\mathrm{CO})_{4}\). (b) Show that the overall \(Q_{\mathrm{c}}\) is the product of the \(Q_{c}\) 's for the individual reactions.

Isopentyl alcohol reacts with pure acetic acid to form isopentyl acetate, the essence of banana oil: $$ \mathrm{C}_{5} \mathrm{H}_{11} \mathrm{OH}+\mathrm{CH}_{3} \mathrm{COOH} \rightleftharpoons \mathrm{CH}_{3} \mathrm{COOC}_{5} \mathrm{H}_{11}+\mathrm{H}_{2} \mathrm{O} $$ A student adds a drying agent to remove \(\mathrm{H}_{2} \mathrm{O}\) and thus increase the yield of banana oil. Is this approach reasonable? Explain.
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