Question

Consider the following equilibrium, for which \(K_{p}=0.0752\) at $$\begin{array}{l} 480{ }^{\circ} \mathrm{C}: \\ \quad 2 \mathrm{Cl}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 4 \mathrm{HCl}(g)+\mathrm{O}_{2}(g) \end{array}$$ (a) What is the value of \(K_{p}\) for the reaction \(4 \mathrm{HCl}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{Cl}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) ?\) (b) What is the value of \(K_{p}\) for the reaction \(\mathrm{Cl}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons\) \(2 \mathrm{HCl}(g)+\frac{1}{2} \mathrm{O}_{2}(g) ?(\mathrm{c})\) What is the value of \(K_{c}\) for the reac- tion in part (b)?

Short answer

The short answers for the given question are as follows: (a) The \(K_p\) value for the reverse reaction is 13.30. (b) The \(K_p\) value for the half-reaction is 0.274. (c) The \(K_c\) value for the reaction in part (b) is 0.0411.

Step-by-step solution

(a) Calculate \(K_{p}\) for the reverse reaction.

To obtain \(K_p\) for the reverse reaction, we simply need to take the reciprocal of the given \(K_p\) value. This is because the reverse reaction corresponds to the inverse of the original reaction. So, for the reaction \(4\mathrm{HCl}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2\mathrm{Cl}_{2}(g)+2\mathrm{H}_{2}\mathrm{O}(g)\), we find the \(K_p\) value as follows: \[K_p' = \frac{1}{K_p} = \frac{1}{0.0752}\]

(a) Calculate the value of the new \(K_{p}\).

By calculating the reciprocal, we get: \[K_p' = 13.30\] So, the \(K_p\) value for the reverse reaction is 13.30.

(b) Calculate \(K_{p}\) for the half-reaction.

For this part, we need to consider the reaction: \(\mathrm{Cl}_{2}(g)+\mathrm{H}_{2}\mathrm{O}(g) \rightleftharpoons 2\mathrm{HCl}(g)+\frac{1}{2}\mathrm{O}_{2}(g)\). This new reaction is obtained by dividing the given reaction by 2. When we halve a reaction, the new \(K_p\) value is found by taking the square root of the original \(K_p\). So, \[K_p'' = \sqrt{K_p} = \sqrt{0.0752}\]

(b) Calculate the value of the new \(K_{p}\).

By calculating the square root, we get: \[K_p'' = 0.274\] Therefore, the \(K_p\) value for the half-reaction is 0.274.

(c) Calculate \(K_{c}\) for the half-reaction.

To find the \(K_c\) value for the half-reaction, we first need the relationship between \(K_p\) and \(K_c\). The relationship is given by: \[K_p = K_c(RT)^{\Delta n}\] where \(R\) is the universal gas constant (0.0821 L·atm/mol·K), \(T\) is the temperature in Kelvin and \(\Delta n\) is the difference in the moles of gases between products and reactants. For the given reaction in part (b), \(\Delta n = (2 - 1 - \frac{1}{2}) = \frac{1}{2}\). The temperature given is \(480{ }^{\circ} \mathrm{C}\) which is equal to \(480 + 273.15 = 753.15 K\). Using this information, we can find the \(K_c\) value for the half-reaction. Firstly, isolate \(K_c\) in the equation: \[K_c = \frac{K_p}{(RT)^{\Delta n}}\]

(c) Calculate the value of \(K_c\).

Now, we plug in the values of \(K_p\), \(R\), \(T\), and \(\Delta n\), and solve for \(K_c\): \[K_c = \frac{0.274}{(0.0821 \times 753.15)^{1/2}}\] After the calculation, we obtain: \[K_c = 0.0411\] Thus, the value of \(K_c\) for the reaction in part (b) is 0.0411.

Key concepts

Kp for Reversible Reactions

Understanding the equilibrium constant expressed in terms of partial pressures (\(K_p\)) is crucial when dealing with gaseous reversible reactions. The value of \(K_p\) is directly tied to the concentrations of the reactants and products at equilibrium. When the reaction reaches the point where the rates of the forward and reverse reactions are equal, the \((K_p)\) for the reaction can be calculated using the partial pressures of the gases involved.

For a given balanced chemical equation, when the direction of the reaction is reversed, the new \((K_p')\) is the reciprocal of the original \(K_p\). This is because reversing the reaction essentially swaps the roles of reactants and products. Illustrating this concept, consider a reaction \((A \rightleftharpoons B)\) with \(K_p = x\), for the reverse reaction \((B \rightleftharpoons A)\), \((K_p') \) would be \((1/x)\), keeping the principle of equilibrium in mind.

Furthermore, if the balanced equation is multiplied or divided by a number, \(K_p\) will be raised to the power of that number. This is evident in a reaction where halving the equation (thus dividing by 2) leads to taking the square root of the original \(K_p\) to obtain the new \(K_p''\). This underscores the importance of stoichiometry in determining equilibrium constants and showcases how changes in the balanced equation affect \(K_p\).

Relationship Between Kp and Kc

The relationship between \(K_p\) and \(K_c\), the equilibrium constants for partial pressures and concentrations, respectively, is a fundamental concept in chemical equilibrium. The two are interrelated by the equation \(K_p = K_c (RT)^{\Delta n}\), where \(R\) is the universal gas constant, \(T\) is the temperature in Kelvin, and \(\Delta n\) is the change in the number of moles of gas across the reaction.

Understanding this relationship is key when converting equilibrium constants from one form to another. The temperature must always be in Kelvin for this equation to work correctly. In this context, \(\Delta n\) presents the net difference in the moles of gaseous products and gaseous reactants. It accounts for the physical basis behind the equilibrium constant expressions, illustrating how changes in volume and pressure influence the equilibrium position.

For instance, in the example given with the reaction at \(480{ }^\circ \text{C}\), we account for the temperature conversion to Kelvin and apply the difference in mole numbers to properly transition between \(K_c\) and \(K_p\). This highlights the need to understand the properties of gases and how they interact under different conditions to accurately describe and predict the behavior of chemical systems at equilibrium.

Calculating Equilibrium Constants

Calculating equilibrium constants, whether \(K_p\) or \(K_c\), involves a systematic approach to analyzing the balanced chemical equation and the conditions of the system at equilibrium. Both \((K_p)\) and \(K_c\) provide a quantitative measure of the position of equilibrium, with their values indicating the extent to which a reaction proceeds.

For \(K_p\), it is calculated based on the partial pressures of the gaseous components. On the other hand, \(K_c\) is derived from the molar concentrations of reactants and products in solution. To compute these constants, one must first ensure the reaction is at equilibrium and that all necessary data, such as partial pressures or concentrations and the temperature, is accurately measured.

In exercises where the calculation of \(K_c\) is required from known \(K_p\) values, the ideal gas law becomes instrumental along with the expression that relates \(K_p\) to \(K_c\). Through these calculations, students can discern the profound effects of reaction conditions, such as temperature and pressure, on the chemical equilibrium and develop a deep understanding of the principles that govern chemical reactions.