Question

At \(1000 \mathrm{~K}, K_{p}=1.85\) for the reaction $$ \mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{SO}_{3}(g) $$ (a) What is the value of \(K_{p}\) for the reaction \(\mathrm{SO}_{3}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) ?\) (b) What is the value of \(K_{p}\) for the reaction \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)\) ? (c) What is the value of \(K_{c}\) for the reaction in part (b)?

Short answer

(a) The value of \(K_{p}\) for the reverse reaction is 0.5405. (b) The value of \(K_{p}\) for the double forward reaction is 3.4225. (c) The value of \(K_{c}\) for the reaction in part (b) is 281.245.

Step-by-step solution

(a) Find the \(K_p\) for the reverse reaction:

To find the \(K_p\) value for the reverse reaction, we need to take the reciprocal of the given \(K_p\). For the reverse reaction \(\mathrm{SO}_{3}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g)\), we have: $$ K_{p(reverse)} = \frac{1}{K_{p(forward)}} $$ $$ K_{p(reverse)} = \frac{1}{1.85} $$

(a) Calculate the \(K_p\) for the reverse reaction:

Now, we can calculate the value for \(K_{p(reverse)}\): $$ K_{p(reverse)} = 0.5405 $$

(b) Find the \(K_p\) for the double of the forward reaction:

For the double reaction \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)\), the equilibrium constant changes according to the following rule: $$ K_{p(2x)} = K_{p}^{n} $$ where n is the factor by which we multiplied the reaction (in this case, n = 2)

(b) Calculate the \(K_p\) for the double forward reaction:

Now, we can calculate the value of \(K_p\) for the double forward reaction: $$ K_{p(2x)} = (1.85)^2 $$ $$ K_{p(2x)} = 3.4225 $$

(c) Find the relation between \(K_p\) and \(K_c\):

We have the following relation between \(K_p\) and \(K_c\): $$ K_p = K_c(RT)^{Δn} $$ where \(Δn\) is the change in moles of gas in the reaction, R is the ideal gas constant, and T is the temperature in Kelvin. For the reaction in part (b), \(Δn = -1\). The given temperature is 1000 K and the ideal gas constant R = 0.0821 L.atm/(K.mol)

(c) Calculate \(K_c\) for the reaction in part (b):

We can now find \(K_c\) for the reaction in part (b) using the relation between \(K_p\) and \(K_c\): $$ K_c = \frac{K_p}{(RT)^{Δn}} $$ $$ K_c = \frac{3.4225}{(0.0821 \times 1000)^{-1}} $$ Now, we can calculate the value of \(K_c\): $$ K_c = 281.245 $$ To summarize, we found: (a) \(K_{p(reverse)} = 0.5405\) (b) \(K_{p(2x)} = 3.4225\) (c) \(K_{c(2x)} = 281.245\)

Key concepts

Chemical Equilibrium

Chemical equilibrium is a fundamental concept in chemistry that occurs when the rate of the forward reaction equals the rate of the reverse reaction in a closed system, leading to a constant concentration of reactants and products over time. It's important to understand that this doesn't mean the reactants and products are in equal concentrations, but rather that their concentrations have stabilized to a certain ratio.

For the given reaction involving sulfur dioxide and oxygen producing sulfur trioxide, the equilibrium constant, denoted as \(K_p\), is given at 1000 K. The equilibrium constant is crucial because it provides us with the ratio of product to reactant concentrations at equilibrium for gas-phase reactions, with the 'p' in \(K_p\) indicating that the concentrations are expressed in terms of partial pressures. When the reaction is reversed, the value of the equilibrium constant also reverses (becomes the reciprocal), and if the equation is multiplied by a factor, the equilibrium constant is raised to the power of that factor, as shown in the textbook solutions.

Le Chatelier's Principle

Le Chatelier's principle helps us predict how a system at equilibrium reacts to external changes. This principle states that if an external change, such as pressure, concentration, or temperature, is applied to a system at equilibrium, the system adjusts itself to minimize the effect of that change.

This principle is pivotal in understanding the behavior of equilibrium systems in response to changes. If we were to increase the pressure on the system given in the exercise, for instance, by adding more \(SO_2(g)\) or \(O_2(g)\), the equilibrium would shift towards the \(SO_3(g)\) to reduce pressure. Conversely, decreasing the concentration of \(SO_3(g)\) would shift the equilibrium towards producing more \(SO_3(g)\) from \(SO_2(g)\) and \(O_2(g)\), according to Le Chatelier's principle.

Reaction Quotient

The reaction quotient, denoted as \(Q\), is a crucial concept that tells us the direction in which a reaction will proceed to reach equilibrium. It involves the same expression as the equilibrium constant but uses the initial concentrations or partial pressures of reactants and products, not those at equilibrium. Upon comparing \(Q\) with \(K_p\), if \(Q < K_p\), the reaction will proceed forwards to reach equilibrium, and if \(Q > K_p\), the reaction will go in the reverse direction.

In the case of the exercise, one would calculate \(Q\) using the initial pressures or concentrations when the system is not at equilibrium. Then by comparing \(Q\) to the calculated \(K_p\) values, we can deduce in which direction the reaction will proceed to achieve equilibrium.

Gas-Phase Reactions

Gas-phase reactions are characterized by reactants and products that are gases, and their equilibrium states can be described using partial pressures. The equilibrium constant for gas-phase reactions is usually expressed in terms of partial pressures and denoted as \(K_p\). \(K_p\) is related to the equilibrium concentrations of the gases, represented by \(K_c\), through the equation \(K_p = K_c(RT)^{\Delta n}\), where \(\Delta n\) is the difference in moles between products and reactants, \(R\) is the ideal gas constant, and \(T\) is the temperature.

In the example provided, the reactions involve gases and the application of the ideal gas law helps us to convert between \(K_p\) and \(K_c\), providing insight into how pressure and concentration are related at equilibrium. The steps illustrate the calculation of the equilibrium constant for the reverse reaction and the reaction with altered stoichiometry, both of which are vital concepts when dealing with gas-phase reactions.