Problem 91 - Chemistry The Central Science

Question

An equilibrium mixture of \(\mathrm{H}_{2}, \mathrm{I}_{2},\) and \(\mathrm{HI}\) at \(458^{\circ} \mathrm{C}\) contains \(0.112 \mathrm{mol} \mathrm{H}_{2}, 0.112 \mathrm{mol} \mathrm{I}_{2},\) and 0.775 \(\mathrm{mol}\) HI in a 5.00 -L. vessel. What are the equilibrium partial pressures when equilibrium is reestablished following the addition of 0.200 \(\mathrm{mol}\) of \(\mathrm{HI}\) ?

Step-by-Step Solution

Answer

After the addition of 0.200 mol HI, the equilibrium partial pressures in the reaction mixture are approximately 0.56 atm for both H2 and I2, and 9.09 atm for HI.

Step 1: Calculate the initial concentrations

Dividing the initial moles of each substance by the volume of the container (5.00 L) will give us the initial concentrations in M (molarity): Initial Molarity of H2 = \(\frac{0.112 \ mol}{5.00 \ L}\) = 0.0224 M Initial Molarity of I2 = \(\frac{0.112 \ mol}{5.00 \ L}\) = 0.0224 M Initial Molarity of HI = \(\frac{0.775 \ mol}{5.00 \ L}\) = 0.155 M

Step 2: Calculate the initial concentrations with the added HI

Now, we need to adjust the initial concentrations by adding the additional amount of HI: New concentration of HI = \(\frac{(0.775 + 0.200) \ mol}{5.00 \ L}\) = \(\frac{0.975 \ mol}{5.00 \ L}\) = 0.195 M

Step 3: Set up the reaction table

The reaction is $$H_2 + I_2 \rightleftharpoons 2HI$$ To determine the equilibrium concentrations, we will set up a table showing the initial concentrations (I), changes (C), and equilibrium concentrations (E) for the reactants and products: Substance: H2 I2 2HI Initial (I): 0.0224 M 0.0224 M 0.195 M Change (C): -x -x +2x Equilibrium (E): 0.0224-x 0.0224-x 0.195+2x

Step 4: Write the equilibrium expression and solve for x

The equilibrium constant expression for the reaction is: \(K_c = \frac{[HI]^2}{[H_2][I_2]}\) Since we are not given the value of Kc, we cannot directly solve the equation. However, we know that the initial conditions were at equilibrium, so we can use the initial concentrations to find Kc: Kc =\(\frac{[0.155]^2}{[0.0224][0.0224]}\) Kc = \(\frac{0.024025}{0.00050176}\) Kc = 47.9 Now that we have the value of Kc, we can use the equilibrium concentrations to find x: \(47.9 = \frac{[0.195+2x]^2}{[(0.0224-x)(0.0224-x)]}\)

Step 5: Solve for x

To solve for x, we now have to solve the above equation. In practice, it might be necessary to use a numerical method, such as the quadratic formula, to find a solution. However, in this case, we can approximate the solution by assuming that x << 0.0224, making the quadratic equation solvable: \(47.9 = \frac{[0.195+2x]^2}{[(0.0224)(0.0224)]}\) Solving for x: x ≈ 0.00926 M

Step 6: Calculate the equilibrium partial pressures

We now have the value of x, which we can use to calculate the equilibrium concentrations: Equilibrium concentration of H2 = 0.0224 - x = 0.01314 M Equilibrium concentration of I2 = 0.0224 - x = 0.01314 M Equilibrium concentration of HI = 0.195 + 2x = 0.21352 M Finally, we can find the equilibrium partial pressures using the ideal gas law: Equilibrium partial pressure of H2 = 0.01314 M \(\times\) 0.08206 L atm/mol K \(\times\) (458 + 273) K = 0.56 atm Equilibrium partial pressure of I2 = 0.01314 M \(\times\) 0.08206 L atm/mol K \(\times\) (458 + 273) K = 0.56 atm Equilibrium partial pressure of HI = 0.21352 M \(\times\) 0.08206 L atm/mol K \(\times\) (458 + 273) K = 9.09 atm Therefore, after the addition of 0.200 mol HI, the equilibrium partial pressures in the reaction mixture are 0.56 atm for both H2 and I2, and 9.09 atm for HI.

Key Concepts

Equilibrium Constant

In a chemical reaction at equilibrium, the equilibrium constant, denoted as \( K_c \), is a measure of the ratio of the concentration of products to reactants, each raised to the power of their stoichiometric coefficients. For the reaction \( \mathrm{H}_2 + \mathrm{I}_2 \rightleftharpoons 2\mathrm{HI} \), the equilibrium expression becomes:

• \( K_c = \frac{[\mathrm{HI}]^2}{[\mathrm{H}_2][\mathrm{I}_2]} \)

The value of \( K_c \) indicates whether products or reactants are favored. A large \( K_c \) value, like 47.9 as calculated in the solution, suggests that at equilibrium, the reaction heavily favors the formation of products or, in this case, more HI is present compared to H2 and I2. Understanding \( K_c \) is crucial because it allows chemists to predict the position of equilibrium and manipulate conditions to maximize product yield.

Partial Pressure

Partial pressure refers to the pressure exerted by individual gas species in a mixture of gases. Each gas in a mixture behaves independently, meaning its partial pressure is proportional to its mole fraction and the total pressure of the system. For our reaction taking place in a closed system at equilibrium, the partial pressures are calculated using the ideal gas law, given that we know the concentrations of each gas.

• \( \text{Partial Pressure of } \mathrm{H}_2 = 0.56 \, \text{atm} \)
• \( \text{Partial Pressure of } \mathrm{I}_2 = 0.56 \, \text{atm} \)
• \( \text{Partial Pressure of } \mathrm{HI} = 9.09 \, \text{atm} \)

These values are pivotal because they help in understanding how the system will behave upon changes in conditions, especially in scenarios involving Le Chatelier's principle.

Le Chatelier's Principle

Le Chatelier's Principle provides insight into how equilibrium systems respond to disturbances. It states that if a balanced system is disturbed, it will shift in the direction that counteracts the disturbance. In our scenario, adding HI causes a temporary imbalance.
According to this principle, since more HI was added, the system shifts towards the left, forming more reactants \( \text{(H}_2 \text{ and I}_2 \text{)} \) until a new equilibrium is achieved. This keeps the \( K_c \) value constant even after HI's addition. Le Chatelier's Principle is vital for predicting how changes like alterations in concentration, pressure, or temperature can impact equilibrium, hence aiding in controlling and optimizing chemical processes.

Reaction Quotients

The reaction quotient, \( Q \), serves as a tool for determining the direction in which a reaction will proceed to reach equilibrium. For the equation \( \mathrm{H}_2 + \mathrm{I}_2 \rightleftharpoons 2\mathrm{HI} \), the reaction quotient \( Q \) is calculated similarly to \( K_c \):

• \( Q = \frac{[\mathrm{HI}]^2}{[\mathrm{H}_2][\mathrm{I}_2]} \)

When a system is not at equilibrium:

• If \( Q < K_c \), the forward reaction is favored, meaning more products will form.
• If \( Q > K_c \), the reverse reaction is favored, indicating more reactants need to form.
• If \( Q = K_c \), the system is at equilibrium.

For our exercise, determining \( Q \) after adding HI helps understand how the system will adjust itself until \( Q \) equals \( K_c \).

Ideal Gas Law

The ideal gas law is a fundamental principle that relates the pressure, volume, temperature, and number of moles of a gas through the equation \( PV = nRT \). It is instrumental in calculating partial pressures in systems like ours where gases are involved.

• \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is the temperature in Kelvin.

In contexts involving chemical equilibrium, like our problem, the ideal gas law helps convert concentrations to partial pressures, especially necessary when temperature is constant. Knowing the partial pressures enables us to apply the equilibrium constant \( K_p \), a gaseous equilibrium version of \( K_c \), crucial for understanding the state and dynamics of gas reactions.