Problem 33 - Chemistry The Central Science

Question

The equilibrium \(2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{NOCl}(g)\) is established at 500 \(\mathrm{K}\) . An equilibrium mixture of the three gases has partial pressures of 0.095 atm, 0.171 atm, and 0.28 atm for \(\mathrm{NO}, \mathrm{Cl}_{2},\) and \(\mathrm{NOCl}\) , respectively. (a) Calculate \(K_{p}\) for this reaction at 500.0 \(\mathrm{K}\) . (b) If the vessel has a volume of 5.00 \(\mathrm{L}\) , calculate \(K_{c}\) at this temperature.

Step-by-Step Solution

Answer

a) The equilibrium constant, \(K_p\), at 500.0 K is 9.128. b) The equilibrium constant, \(K_c\), at 500.0 K is 35.01.

Step 1: Write the expression for Kp

We can write the expression for Kp of the given reaction as: \[K_p = \frac{P_{NOCl}^2}{P_{NO}^2 \times P_{Cl_2}}\] where \(P_{NOCl}, P_{NO},\text{ and } P_{Cl_2}\) are the partial pressures of the corresponding gases at equilibrium.

Step 2: Calculate Kp using given pressures

Now we can use the given partial pressures at equilibrium to calculate Kp at 500 K: \[K_p = \frac{(0.28)^2}{(0.095)^2 \times 0.171} \] Using a calculator, we find that: \[K_p = 9.128\]

Step 3: Calculate moles of each species at equilibrium

We are given the volume of the vessel is 5.00 L, and the equilibrium partial pressures. To find the moles of each species at equilibrium, we can use the Ideal Gas Law formula \(PV = nRT\), rearranged as: \[n = \frac{PV}{RT}\] Given that the temperature is 500.0 K and the ideal gas constant is R = 0.0821 atm L mol\(^{-1}\) K\(^{-1}\), we can find the number of moles for each species at equilibrium: \[n_{NO} = \frac{(0.095)(5.00)}{(0.0821)(500)} \Rightarrow n_{NO} = 0.0115\text{ mol}\] \[n_{Cl_2} = \frac{(0.171)(5.00)}{(0.0821)(500)} \Rightarrow n_{Cl_2} = 0.0208\text{ mol}\] \[n_{NOCl} = \frac{(0.28)(5.00)}{(0.0821)(500)} \Rightarrow n_{NOCl} = 0.0340\text{ mol}\]

Step 4: Write the expression for Kc

Now we will write the equilibrium constant, Kc, expression for the given reaction: \[K_c = \frac{[NOCl]^2}{[NO]^2 \times [Cl_2]}\] where \([NOCl], [NO],\text{ and } [Cl_2]\) represent the concentration (in mol L\(^{-1}\)) of the corresponding gases at equilibrium.

Step 5: Calculate Kc using the equilibrium moles

With the equilibrium moles and the volume of 5.00 L, we can calculate the equilibrium concentrations: \[[NO] = \frac{n_{NO}}{V} = \frac{0.0115}{5.00} = 0.0023\text{ mol L}^{-1}\] \[[Cl_2] = \frac{n_{Cl_2}}{V} = \frac{0.0208}{5.00} = 0.00416\text{ mol L}^{-1}\] \[[NOCl] = \frac{n_{NOCl}}{V} = \frac{0.0340}{5.00} = 0.0068\text{ mol L}^{-1}\] Now, we can plug these values into the Kc expression: \[K_c = \frac{(0.0068)^2}{(0.0023)^2 \times 0.00416}\] Using a calculator, we find that: \[K_c = 35.01\] To summarize: a) The equilibrium constant, Kp, at 500.0 K is 9.128. b) The equilibrium constant, Kc, at 500.0 K is 35.01.

Key Concepts

Ideal Gas Law

The Ideal Gas Law is an essential equation in chemistry that relates the pressure, volume, temperature, and number of moles of a gas. Written as \(PV = nRT\), where \(P\) is pressure, \(V\) is volume, \(n\) is the number of moles, \(R\) is the ideal gas constant, and \(T\) is temperature.

To find the number of moles of a gas, rearrange the equation to \(n = \frac{PV}{RT}\). This is particularly useful when you're given pressure, volume, and temperature, as in this exercise.

• \(P\) is in atm
• \(V\) in liters
• \(T\) in Kelvin
• \(R\) typically 0.0821 atm L mol\(^{-1}\) K\(^{-1}\)

This law helps us convert partial pressures to moles, a step critical when moving from \(K_p\) to \(K_c\).

Partial Pressure

Partial pressure refers to the pressure a single gas in a mixture would exert if it occupied the entire volume by itself. It is vital in calculating equilibrium constant \(K_p\) for reactions involving gases.

The equation to calculate \(K_p\) is a reflection of these partial pressures. For the reaction:

• \(2 \mathrm{NO} + \mathrm{Cl}_2 \rightleftharpoons 2 \mathrm{NOCl}\)

The expression becomes:\[K_p = \frac{P_{NOCl}^2}{P_{NO}^2 \times P_{Cl_2}}\]Plug in the given pressures to find \(K_p\). This process gives insight into the equilibrium state under given conditions.

Chemical Equilibrium

Chemical equilibrium occurs when the rates of the forward and reverse reactions are equal, leading to constant concentrations of reactants and products. This steady state allows us to define the equilibrium constant \(K\), which quantifies the position of equilibrium.

\(K_p\) relates to gaseous reactions using partial pressures, whereas \(K_c\) uses concentrations. Understanding these constants helps in predicting reaction behavior under various conditions, such as temperature changes.

• At equilibrium, the forward and reverse reaction rates are equal.
• Concentrations and pressures remain unchanged.

Knowing this enables you to calculate \(K_p\) and \(K_c\), facilitating predictions of reaction outcomes.

Reaction Quotient

The reaction quotient \(Q\) is a snapshot of a system's condition before it reaches equilibrium. It is calculated similarly to the equilibrium constant \(K\) but using initial concentrations or pressures instead of equilibrium values.

\(Q\) helps predict the direction a reaction must proceed to achieve equilibrium:

• If \(Q < K\), the forward reaction is favored.
• If \(Q > K\), the reverse reaction is favored.
• If \(Q = K\), the system is already at equilibrium.

Using \(Q\) is key in evaluating how changes in conditions affect equilibrium, providing insights into reaction dynamics and necessary adjustments.