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Chapter 14: Problem 67

At \(12,000 \mathrm{~K}\) and 6 atm, \(1 \mathrm{kmol}\) of \(\mathrm{N}\) ionizes to form an equilibrium mixture of \(\mathrm{N}^{+} \mathrm{N}^{+}\), and \(\mathrm{e}^{-}\)in which the amount of \(\mathrm{N}\) present is \(0.95 \mathrm{kmol}\). Determine the ionizationequilibrium constant at this temperature for \(\mathrm{N} \rightleftarrows \mathrm{N}^{+}+\mathrm{e}^{-}\).

Short Answer

Expert verified
The ionization equilibrium constant \( K_p \) for \( \text{N} \rightleftarrows \text{N}^+ + e^- \) is 0.01503 atm.

Step by step solution

01

Determine the amount of ionized nitrogen (N+)

Given that the total initial amount of \(\text{N}\) is 1 kmol and the remaining amount of \(\text{N}\) after ionization is 0.95 kmol, the amount of ionized \(\text{N}^+\) can be calculated as follows:$$n_{N^+} = 1 \text{ kmol} - 0.95 \text{ kmol} = 0.05 \text{ kmol}$$
02

Amount of electrons (e-)

Each ionized nitrogen (N) atom produces one electron (e-), so the amount of electrons produced will be equal to the amount of ionized nitrogen atoms:$$n_{e^-} = 0.05 \text{ kmol} $$
03

Calculate the mole fraction

The mole fraction for each component is calculated based on their amount and the total mixture:Total amount of moles = \(n_{\text{N}} + n_{\text{N}^+} + n_{e^-} = 0.95 \text{ kmol} + 0.05 \text{ kmol} + 0.05 \text{ kmol} = 1.05 \text{ kmol} \)\(X_{\text{N}} = \frac{0.95}{1.05} = 0.9048\)\space\(X_{\text{N}^+} = \frac{0.05}{1.05} = 0.0476\)\space\(X_{e^-} = \frac{0.05}{1.05} = 0.0476\)
04

Calculate the partial pressures

At a total pressure of 6 atm, the partial pressures can be calculated as follows:\(P_{\text{N}} = X_{\text{N}} \times 6 \text{ atm} = 0.9048 \times 6 \text{ atm} = 5.4288 \text{ atm}\)\(P_{\text{N}^+} = X_{\text{N}^+} \times 6 \text{ atm} = 0.0476 \times 6 \text{ atm} = 0.2856 \text{ atm}\)\(P_{e^-} = X_{e^-} \times 6 \text{ atm} = 0.0476 \times 6 \text{ atm} = 0.2856 \text{ atm}\)
05

Use the ionization equilibrium equation

The ionization equilibrium constant for the reaction \( \text{N} \rightleftarrows \text{N}^+ + e^- \) is defined as:\[ K_p = \frac{P_{\text{N}^+} \times P_{e^-}}{P_{\text{N}}} \]Substitute the partial pressures calculated in the previous step:\[ K_p = \frac{0.2856 \text{ atm} \times 0.2856 \text{ atm}}{5.4288 \text{ atm}} \]\[ K_p = \frac{0.08155296 \text{ atm}^2}{5.4288 \text{ atm}} = 0.01503 \text{ atm} \]

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Key Concepts

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

Ionization Constant
The ionization constant, often denoted as \( K_p \), quantifies the degree to which a substance ionizes in a reaction. In the context of the given exercise, the ionization constant for the reaction \( \text{N} \rightleftharpoons \text{N}^+ + e^- \) at equilibrium can be found using the equilibrium partial pressures of the substances involved.

To find \( K_p \, \),

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Most popular questions from this chapter

Consider the reactions 1\. \(\mathrm{CO}_{2}+\mathrm{H}_{2} \rightleftarrows \mathrm{CO}+\mathrm{H}_{2} \mathrm{O}\) 2\. \(\mathrm{CO}_{2} \rightleftarrows \mathrm{CO}+\frac{1}{2} \mathrm{O}_{2}\) 3\. \(\mathrm{H}_{2} \mathrm{O} \rightleftarrows \mathrm{H}_{2}+\frac{1}{2} \mathrm{O}_{2}\) (a) Show that \(K_{1}=K_{2} / K_{3}\). (b) Evaluate \(\log _{10} K_{1}\) at \(298 \mathrm{~K}, 1\) atm using the expression from part (a), together with \(\log _{10} K\) data from Table A-27. (c) Check the value for \(\log _{10} K_{1}\) obtained in part (b) by applying Eq. 14.31 to reaction 1 .

Derive an expression for estimating the pressure at which graphite and diamond exist in equilibrium at \(25^{\circ} \mathrm{C}\) in terms of the specific volume, specific Gibbs function, and isothermal compressibility of each phase at \(25^{\circ} \mathrm{C}, 1 \mathrm{~atm}\). Discuss.

An equimolar mixture of carbon monoxide and water vapor at \(200^{\circ} \mathrm{F}, 1 \mathrm{~atm}\) enters a reactor operating at steady state. An equilibrium mixture of \(\mathrm{CO}_{2}, \mathrm{CO}, \mathrm{H}_{2} \mathrm{O}\) (g), and \(\mathrm{H}_{2}\) exits at \(2240^{\circ} \mathrm{F}, 1\) atm. Determine the heat transfer between the reactor and its surroundings, in Btu per lbmol of \(\mathrm{CO}\) entering. Neglect kinetic and potential energy effects.

Carbon dioxide \(\left(\mathrm{CO}_{2}\right)\) and oxygen \(\left(\mathrm{O}_{2}\right)\) in a \(1: 2\) molar ratio enter a reactor operating at steady state in separate streams at \(1 \mathrm{~atm}, 127^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}, 277^{\circ} \mathrm{C}\), respectively. An equilibrium mixture of \(\mathrm{CO}_{2}, \mathrm{CO}\), and \(\mathrm{O}_{2}\) exits at \(1 \mathrm{~atm}\). If the mole fraction of \(\mathrm{CO}\) in the exiting mixture is 0.1, determine the rate of heat transfer from the reactor, in \(\mathrm{kJ}\) per \(\mathrm{kmol}\) of \(\mathrm{CO}_{2}\) entering. Ignore kinetic and potential energy effects.

An equimolar mixture of \(\mathrm{CO}\) and \(\mathrm{O}_{2}\) reacts to form an equilibrium mixture of \(\mathrm{CO}_{2}, \mathrm{CO}\), and \(\mathrm{O}_{2}\) at \(3000 \mathrm{~K}\). Determine the effect of pressure on the composition of the equilibrium mixture. Will lowering the pressure while keeping the temperature fixed increase or decrease the amount of \(\mathrm{CO}_{2}\) present? Explain.
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