Question

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

Short answer

(a) The value of \(K_p\) for the reversed reaction is 13.3. (b) The value of \(K_p\) for the halved reaction is 0.274. (c) The value of \(K_c\) for the halved reaction is \(3.58 \times 10^{-3}\).

Step-by-step solution

Determine the equilibrium constant for the reversed reaction#a.

The given reaction is: \(2 Cl_{2}(g) + 2H_{2}O(g) \rightleftharpoons 4 HCl(g) + O_{2}(g)\) with \(K_{p} = 0.0752\) We can find Kp for the reverse reaction by taking the inverse of the initial reaction's Kp, as the concentrations of reactants and products are reversed. The reverse reaction is: \(4 HCl(g) + O_{2}(g) \rightleftharpoons 2Cl_{2}(g) + 2H_{2}O(g)\) So, for the reverse reaction: \(K_{p}^{'} = \frac{1}{K_{p}} = \frac{1}{0.0752} = 13.3\) Answer (a): The value of Kp for the reversed reaction is 13.3.

Determine the equilibrium constant for the halved reaction#b.

Now, let's consider the halved reaction: \(Cl_{2}(g) + H_{2}O(g) \rightleftharpoons 2 HCl(g) + \frac{1}{2} O_{2}(g)\) Here, the reaction is half of the provided reaction, so we need to find Kp for this new reaction. To do so, we can raise the Kp of the initial reaction to the power of the fraction that we are dividing the reaction by: \(K_{p, half} = K_{p}^{(1/2)} = (0.0752)^{(1/2)} = 0.274\) Answer (b): The value of Kp for the halved reaction is 0.274.

Convert Kp to Kc for the halved reaction#c.

Kp and Kc are related through the ideal gas law with the equation: \(K_{p} = K_{c}(RT)^{\Delta n}\), where R is the gas constant, T is the temperature in Kelvin, and Δn is the change in moles of gases between products and reactants. For the halved reaction, we have Δn = (2 + 1/2) - (1 + 1) = 0.5. The temperature provided is 480°C, which we should convert to Kelvin: \(T = 480 + 273.15 = 753.15 K\) Now, we can find Kc for the halved reaction: \(K_{c} = \frac{K_{p}}{(RT)^{\Delta n}} = \frac{0.274}{(0.0821 \times 753.15)^{0.5}} = 3.58 \times 10^{-3}\) Answer (c): The value of Kc for the halved reaction is \(3.58 \times 10^{-3}\).

Key concepts

Chemical Equilibrium

In chemistry, chemical equilibrium is a state where the concentrations of reactants and products remain constant over time because the rate of the forward reaction equals the rate of the reverse reaction. This doesn't mean the chemical reactions stop occurring; rather, they continue at a steady rate such that there is no net change in the concentrations of the chemical species involved. At equilibrium, the system's properties like concentration, color, pressure, and density become unchanging because the forward and backward reactions are in perfect balance.

At the molecular level, this dynamic balance entails constant interchange between reactants forming products and products reconverting into reactants. Simply, it's like a crowded dance floor where the number of people entering and leaving the floor is the same at any given time. In the context of our problem, for the reaction involving chlorine, water, hydrochloric acid, and oxygen the dance continues with participants (molecules) switching partners (reacting) but with the overall number remaining the same over time.

Kp to Kc Conversion

The conversion from the equilibrium constant expressed in terms of partial pressures \( K_{p} \) to the equilibrium constant expressed in terms of concentrations \( K_{c} \) is an important concept in equilibrium chemistry. This conversion uses the ideal gas law and the relationship between \( K_{p} \) and \( K_{c} \) can be described by the equation \( K_{p} = K_{c} \times (RT)^{\Delta n} \), 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 gases when moving from reactants to products.

For students, this can be envisioned like a currency exchange between two different countries' monies—where \( K_{p} \) and \( K_{c} \) are different 'currencies' and the gas constant \( R \) and temperature \( T \) define the 'exchange rate'. \( \Delta n \) tells us the 'transaction fee' based on how the number of participants (moles of gas) changes during the reaction.

Equilibrium Constant for Reverse Reactions

Understanding the equilibrium constant for reverse reactions is crucial. Essentially, the equilibrium constant \( K \) for a reverse reaction is simply the reciprocal of the constant for the forward reaction. So, if the equilibrium constant for a forward reaction is \( K_f \), then for the reverse reaction, it would be \( K_r = \frac{1}{K_f} \).

Think of it as flipping a video clip; everything happens in the exact reverse order. As seen with our chlorine and water reaction, flipping the reaction equals inverting the equilibrium constant. This mathematical relationship is like a seesaw, where one side goes down (the forward reaction), the other goes up (the reverse reaction), all with a fulcrum of one, representing the balance point between the two reaction directions.

Equilibrium Constant for Halved Reactions

Halving a reaction and its impact on the equilibrium constant \( K \) may seem less intuitive. When a given balanced chemical equation is halved, the equilibrium constant for the halved reaction is the initial equilibrium constant raised to the power of 1/2. In our exercise, the initial reaction is split into two equal parts, which implies a square root operation on \( K_p \).

This action of halving a reaction is akin to dividing a pie into smaller portions—the size of the entire pie doesn't change, but the size of each piece does, and similarly, while the nature of equilibrium doesn't alter, the numerical value of the constant does. Therefore, to find the new equilibrium constant for the halved reaction, we must adjust the initial equilibrium constant with the power of 1/2 to reflect this change.