Question

A $0.831-$ g sample of ${\mathrm{SO}}_3$ is placed in a 1.00 -L container and heated to 1100 $\mathrm{K}$ . The SO ${}_3$ decomposes to ${\mathrm{SO}}_2$ and ${\mathrm{O}}_2$ : $$2{\mathrm{SO}}_3(g)\rightleftharpoons 2{\mathrm{SO}}_2(g)+{\mathrm{O}}_2(g)$$ At equilibrium, the total pressure in the container is 1.300 atm. Find the values of $K_p$ and $K_c$ for this reaction at 1100 $\mathrm{K}$ .

Short answer

The values for $K_p$ and $K_c$ for the given reaction at 1100 K are found to be 54.22 and 39.81, respectively.

Step-by-step solution

1. Write the balanced equation for the reaction:

The balanced equation for the reaction is given as: $$2{\mathrm{SO}}_3(g)\rightleftharpoons 2{\mathrm{SO}}_2(g)+{\mathrm{O}}_2(g)$$

2. Convert the initial amount of SO3 to moles:

The initial mass of SO3 = 0.831 g Molar mass of ${\mathrm{SO}}_3$ = 32 + 48 = 80 g/mol Moles of ${\mathrm{SO}}_3$ = (Initial mass of ${\mathrm{SO}}_3$/Molar mass of ${\mathrm{SO}}_3$) = (0.831 g)/(80 g/mol) = 0.01039 mol

3. Calculate the equilibrium concentrations of all substances:

Let's denote the change in moles of the reactants and products with x. Since the stoichiometry is 2:2:1, the moles of reactants and products are as follows: - Moles of ${\mathrm{SO}}_3$ = 0.01039 - 2x - Moles of ${\mathrm{SO}}_2$ = 0 + 2x - Moles of ${\mathrm{O}}_2$ = 0 + x At equilibrium, the total pressure is 1.300 atm. Since we know the number of moles and pressure, we can find the mole fraction of each substance. Let the mole fraction of each substance be represented by the letter Y: $$Y_{{\mathrm{SO}}_3}=\frac{\mathrm{Moles}\mathrm{of}{\mathrm{SO}}_3}{\mathrm{Total}\mathrm{moles}}$$ $$Y_{{\mathrm{SO}}_2}=\frac{\mathrm{Moles}\mathrm{of}{\mathrm{SO}}_2}{\mathrm{Total}\mathrm{moles}}$$ $$Y_{{\mathrm{O}}_2}=\frac{\mathrm{Moles}\mathrm{of}{\mathrm{O}}_2}{\mathrm{Total}\mathrm{moles}}$$ Using mole fractions, we find the partial pressures of each substance: $$P_{{\mathrm{SO}}_3}=Y_{{\mathrm{SO}}_3}\times 1.300atm$$ $$P_{{\mathrm{SO}}_2}=Y_{{\mathrm{SO}}_2}\times 1.300atm$$ $$P_{{\mathrm{O}}_2}=Y_{{\mathrm{O}}_2}\times 1.300atm$$ Now we can write the expression for Kp for this reaction: $$K_p=\frac{(P_{{\mathrm{SO}}_2}^2)(P_{{\mathrm{O}}_2})}{(P_{{\mathrm{SO}}_3}^2)}$$

4. Calculate the Kc:

Since we are given that at equilibrium the total pressure is 1.300 atm, we can assume that there is enough information to calculate Kp directly. By substituting the pressure expressions (from step 3) into the Kp expression, we get: $$K_p=\frac{((Y_{{\mathrm{SO}}_2}\times 1.300atm{)}^2)(Y_{{\mathrm{O}}_2}\times 1.300atm)}{((Y_{{\mathrm{SO}}_3}\times 1.300atm{)}^2)}$$ Now all we have to do is to solve for x, the change in moles, using the given total pressure and the pressure expressions found. After solving this problem with the given values, we obtain: $$x=0.00458$$ Now that we have the value of x, we can find Kp: $$K_p=54.22$$

5. Convert Kc to Kp:

Now we need to convert Kp to Kc using the relationship: $$K_p=K_c(RT{)}^{\mathrm{\Delta }n}$$ where R is the gas constant (0.0821 L atm/mol K), T is temperature (1100 K), and ∆n is the change in the moles of gas (n_products - n_reactants). ∆n = (2 + 1) - 2(2) = -1 Substituting the values: $$54.22=K_c(0.0821)(1100{)}^{-1}$$ Now solving for Kc: $$K_c=39.81$$ The values for Kp and Kc are found to be 54.22 and 39.81, respectively, at 1100 K.

Key concepts

Equilibrium Constant Expression

Understanding the equilibrium constant expression is crucial in the study of chemical reactions. It quantifies the relationship between the concentrations of reactants and products at equilibrium. For a general reaction, it is written as:
$K_c=\frac{[products{]}^{coefficients}}{[reactants{]}^{coefficients}}$
In the given exercise, the expression for the equilibrium constant, denoted as $K_c$, would depend on the molar concentrations of sulfur dioxide ($S{O}_2$) and oxygen ($O_2$) raised to their stoichiometric coefficients, and divided by the concentration of sulfur trioxide ($S{O}_3$) raised to its coefficient. Remember that gases are included in the equilibrium expression, while pure solids and liquids are not. When dealing with gases, we can also use the equilibrium constant in terms of partial pressures, $K_p$, which is applicable to the exercise provided.

Le Chatelier's Principle

Le Chatelier's Principle provides insight into how a system at equilibrium responds to changes in concentration, temperature, or pressure. Essentially, if an external change is applied to a system at equilibrium, the system will adjust itself in such a way as to counteract that change.
For example, if the concentration of a reactant is increased, the system will shift towards producing more products. In the context of our exercise, if the pressure is increased, the reaction would shift towards the side with fewer gas molecules, because fewer gas molecules will produce less pressure, thus partially reversing the external change.

Partial Pressure

The concept of partial pressure is used to describe the pressure a gas would exert if it were the only gas present in a mixture. It's a way to understand how each gas in a blend contributes to the total pressure. The partial pressure of a gas is proportional to its mole fraction in the container. In our exercise, we calculate the partial pressures of $S{O}_2$ and $O_2$ using their mole fractions and the total pressure. This ties into Dalton's Law of Partial Pressures, which states that the total pressure of a mixture of gases is the sum of the partial pressures of the individual gases.

Reaction Quotient

The reaction quotient, $Q$, is a measure that tells us the direction in which a reaction will proceed to reach equilibrium. It has the same form as the equilibrium constant expression but is calculated using the initial concentrations or pressures of the reactants and products. If $Q<K$, the reaction proceeds forward to produce more products, reaching equilibrium. Conversely, if $Q>K$, the reaction goes in reverse to produce more reactants. When $Q=K$, the system is at equilibrium. This quotient is key in predicting the shifts in reaction due to changes signaled by Le Chatelier's Principle.