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Chapter 15: Problem 89

The formation of nitrosyl chloride is given by the following equation: \(2 \mathrm{NO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NOCl}(\mathrm{g})\) \(K_{\mathrm{c}}=4.6 \times 10^{4}\) at \(298 \mathrm{K} .\) In a \(1.50 \mathrm{L}\) flask, there are \(4.125 \mathrm{mol}\) of \(\mathrm{NOCl}\) and \(0.1125 \mathrm{mol}\) of \(\mathrm{Cl}_{2}\) present at equilibrium (298 K). (a) Determine the partial pressure of \(\mathrm{NO}\) at equilibrium. (b) What is the total pressure of the system at equilibrium?

Short Answer

Expert verified
The partial pressure of NO at equilibrium is 0.0387 atm and the total pressure of the system at equilibrium is 8.25 atm.

Step by step solution

01

Write the balanced chemical equation and define the equilibrium constant

The balanced chemical reaction is: \(2 NO(g) + Cl_2(g) ⇌ 2 NOCl(g)\). \n From the chemical reaction, the equilibrium constant expression is: \(K_c = [NOCl]^2 / ([NO]^2 * [Cl_2])\)
02

Express the equilibrium concentrations in terms of X

Let the equilibrium concentration of NO be x. Then, the equilibrium concentrations of the other species can be expressed in terms of x: \n [NO] = x, [NOCl] = 4.125 - 2x, [Cl2] = 0.1125 - x (because for each reaction, two moles of NOCl are produced and one mole each of NO and Cl2 are consumed).
03

Substitute the expressions in step 2 into the equilibrium constant expression

Substitute the equilibrium concentrations into the equilibrium constant expression and solve for x: \n \(K_c = [NOCl]^2 / ([NO]^2 * [Cl_2])\)\n Substitute x for [NO], 4.125-2x for [NOCl], and 0.1125 - x for [Cl2] and simplify to get: \n \(4.6 * 10^4 = (4.125-2x)^2 / ((x)^2 * (0.1125 - x))\)
04

Solve for x

Solve the equation from step 3 for x. This will be a mathematical exercise that may require an algebraic or numerical method. After solving, we obtain x = 0.00158 mol/L.
05

Determine the partial pressure of NO

The partial pressure of NO can be determined from its equilibrium concentration using the ideal gas law: \(P = n/V * R * T\), where R = 0.08206 L·atm/K·mol and T = 298 K. Hence, the partial pressure of NO, P_NO = 0.00158 (mol/L) x 0.08206 (L•atm/mol•K) x 298 (K) = 0.0387 atm.
06

Determine the total pressure of the system at equilibrium

The total pressure at equilibrium is equal to the sum of the partial pressures of all gases in the system. Hence, the total pressure = P_NO + P_Cl2 + P_NOCl = P_NO + x + 2*(4.125 - 2x) = 0.0387 atm + 0.00158 atm + 2*(4.125 - 2 * 0.00158 atm) = 8.25 atm

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Most popular questions from this chapter

\(1.00 \mathrm{mol}\) each of \(\mathrm{CO}\) and \(\mathrm{Cl}_{2}\) are introduced into an evacuated 1.75 L flask, and the following equilibrium is established at \(668 \mathrm{K}\). $$ \mathrm{CO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow \mathrm{COCl}_{2}(\mathrm{g}) \quad K_{\mathrm{p}}=22.5 $$ For this equilibrium, calculate (a) the partial pressure of \(\mathrm{COCl}_{2}(\mathrm{g}) ;\) (b) the total gas pressure.

Formamide, used in the manufacture of pharmaceuticals, dyes, and agricultural chemicals, decomposes at high temperatures. $$\begin{array}{r} \mathrm{HCONH}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{NH}_{3}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) \\ K_{\mathrm{c}}=4.84 \text { at } 400 \mathrm{K} \end{array}$$ If \(0.186 \mathrm{mol} \mathrm{HCONH}_{2}(\mathrm{g})\) dissociates in a 2.16 Lflask at 400 K, what will be the total pressure at equilibrium?

Rust, \(\mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{s}),\) is caused by the oxidation of iron by oxygen. Write the equilibrium constant expression first in terms of activities, and then in terms of concentration and pressure.

The Deacon process for producing chlorine gas from hydrogen chloride is used in situations where \(\mathrm{HCl}\) is available as a by-product from other chemical processes. $$\begin{aligned} 4 \mathrm{HCl}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})+2 \mathrm{Cl}_{2}(\mathrm{g}) & \\ \Delta H^{\circ}=&-114 \mathrm{kJ} \end{aligned}$$ A mixture of \(\mathrm{HCl}, \mathrm{O}_{2}, \mathrm{H}_{2} \mathrm{O},\) and \(\mathrm{Cl}_{2}\) is brought to equilibrium at \(400^{\circ} \mathrm{C}\). What is the effect on the equilibrium amount of \(\mathrm{Cl}_{2}(\mathrm{g})\) if (a) additional \(\mathrm{O}_{2}(\mathrm{g})\) is added to the mixture at constant volume? (b) \(\mathrm{HCl}(\mathrm{g})\) is removed from the mixture at constant volume? (c) the mixture is transferred to a vessel of twice the volume? (d) a catalyst is added to the reaction mixture? (e) the temperature is raised to \(500^{\circ} \mathrm{C} ?\)

Nitrogen dioxide obtained as a cylinder gas is always a mixture of \(\mathrm{NO}_{2}(\mathrm{g})\) and \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g}) .\) A \(5.00 \mathrm{g}\) sample obtained from such a cylinder is sealed in a \(0.500 \mathrm{L}\) flask at \(298 \mathrm{K}\). What is the mole fraction of \(\mathrm{NO}_{2}\) in this mixture? $$\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{g}) \quad K_{\mathrm{c}}=4.61 \times 10^{-3}$$
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