Question

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

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

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: \(K_c = [NOCl]^2 / ([NO]^2 * [Cl_2])\)

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).

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 \(4.6 * 10^4 = (4.125-2x)^2 / ((x)^2 * (0.1125 - x))\)

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.

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.

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

Key concepts

Equilibrium Constant

The equilibrium constant, represented as \(K_c\) for reactions involving concentrations, is a crucial concept in chemical equilibrium. It shows the ratio of the concentrations of products to reactants at equilibrium. These concentrations are raised to the power of their coefficients from the balanced chemical equation. For the reaction \(2 \mathrm{NO}(\mathrm{g}) + \mathrm{Cl}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NOCl}(\mathrm{g})\), the expression for \(K_c\) will be \(K_c = \frac{[\mathrm{NOCl}]^2}{[\mathrm{NO}]^2 [\mathrm{Cl}_{2}]}\). A high \(K_c\) value, like \(4.6 \times 10^4\), indicates that at equilibrium, the concentration of products is significantly higher than that of reactants, meaning the reaction heavily favors the formation of products.
Understanding the equilibrium constant allows you to predict the direction of the reaction under different conditions, thereby giving insight into reaction yields and the position of equilibrium.

Partial Pressure

Partial pressure refers to the pressure exerted by a single gas in a mixture of gases. It is directly linked to the mole fraction of the gas in the mixture and the total pressure of the mixture. In equilibrium problems, like with the reaction \(2 \mathrm{NO}(\mathrm{g}) + \mathrm{Cl}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NOCl}(\mathrm{g})\), finding partial pressures helps us understand the dynamics and distribution of gases at equilibrium.
Each gas component contributes to the total pressure based on its concentration or amount. The partial pressure of \(\mathrm{NO}\) was calculated using the Ideal Gas Law, resulting in \(0.0387\) atm. This exemplifies how the partial pressure is proportional to the number of moles and the volume of the system. Being able to calculate partial pressures not only supports solving equilibrium problems but is also vital in determining reaction extents and conditions for gas mixtures.

Ideal Gas Law

The Ideal Gas Law is an equation of state for gases that relates pressure \(P\), volume \(V\), temperature \(T\), and number of moles \(n\). It is represented by \(PV = nRT\), where \(R\) is the ideal gas constant \(0.08206\, \text{L atm/mol K}\). This equation assumes gases are "ideal," meaning their particles do not interact and occupy no volume.
In exercises involving equilibrium, the Ideal Gas Law becomes a handy tool for converting between moles and pressure, as each state's parameters are interdependent. For instance, in the original exercise, it helped calculate the partial pressures required for further equilibrium analysis.
Applying the Ideal Gas Law teaches students to comprehend how gases behave under various conditions and how changes in one parameter affect the others within a gas system.