Problem 62 - Chemistry The Molecular Science

Question

Consider the equilibrium $$ \mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g}) $$ At \(745 \mathrm{~K}\) the equilibrium constant \(K_{\mathrm{c}}=50.0 .\) Suppose that \(0.75 \mathrm{~mol} \mathrm{HI}(\mathrm{g}), 0.025 \mathrm{~mol} \mathrm{H}_{2}(\mathrm{~g})\), and \(0.025 \mathrm{~mol}\) \(\mathrm{I}_{2}(\mathrm{~g})\) are placed into a sealed 20.0 - \(\mathrm{L}\) flask and heated to \(745 \mathrm{~K}\) (a) Is the system at equilibrium? (b) If not, in which direction must the reaction proceed to reach equilibrium? (c) Calculate the equilibrium concentrations of all three substances.

Step-by-Step Solution

Answer

(a) No, it's not at equilibrium. (b) Reaction proceeds to the left. (c) Equilibrium concentrations: [HI] = 0.03543 mol/L; [H2] = 0.002285 mol/L; [I2] = 0.002285 mol/L.

Step 1: Calculate Initial Concentrations

We have given amounts of gases in a 20.0 L flask. Calculate the initial concentrations using the formula \( C = \frac{n}{V} \), where \( n \) is the number of moles and \( V \) is the volume.- For HI: \( C_{\mathrm{HI}} = \frac{0.75 \ \mathrm{mol}}{20.0 \ \mathrm{L}} = 0.0375 \ \mathrm{mol/L} \)- For \( \mathrm{H}_2 \): \( C_{\mathrm{H}_2} = \frac{0.025 \ \mathrm{mol}}{20.0 \ \mathrm{L}} = 0.00125 \ \mathrm{mol/L} \)- For \( \mathrm{I}_2 \): \( C_{\mathrm{I}_2} = \frac{0.025 \ \mathrm{mol}}{20.0 \ \mathrm{L}} = 0.00125 \ \mathrm{mol/L} \)

Step 2: Calculate Reaction Quotient, Q

Use the initial concentrations to calculate the reaction quotient, \( Q \), which is a snapshot measure of the concentration ratio of products to reactants.\[ Q_c = \frac{[\mathrm{HI}]^2}{[\mathrm{H}_2][\mathrm{I}_2]} \]Plug in the initial concentrations:\[ Q_c = \frac{(0.0375)^2}{(0.00125)(0.00125)} = \frac{0.00140625}{0.0000015625} = 900 \]

Step 3: Compare Q with K_c

The calculated \( Q_c \) is 900, while the \( K_c \) is 50.0.- Since \( Q_c > K_c \), the reaction quotient is greater than the equilibrium constant. - This indicates that the system has more products than needed for equilibrium.

Step 4: Determine Reaction Direction

Since \( Q_c > K_c \), the reaction will proceed in the direction that reduces the amount of products; therefore, it will shift to the left, converting \( \mathrm{HI} \) back into \( \mathrm{H}_2 \) and \( \mathrm{I}_2 \).

Step 5: Define Changes for Equilibrium

Let \( x \) be the change in concentration of \( \mathrm{H}_2 \) and \( \mathrm{I}_2 \) as the reaction proceeds towards equilibrium.- The change for \( \mathrm{H}_2 \) and \( \mathrm{I}_2 \) will be \(+x\), while for \( \mathrm{HI} \) it will be \(-2x\).Express the equilibrium concentrations:- \( [\mathrm{HI}] = 0.0375 - 2x \)- \( [\mathrm{H}_2] = 0.00125 + x \)- \( [\mathrm{I}_2] = 0.00125 + x \)

Step 6: Plug into Equilibrium Constant Expression

Use the equilibrium concentrations in the \( K_c \) expression:\[ K_c = \frac{[\mathrm{HI}]^2}{[\mathrm{H}_2][\mathrm{I}_2]} = 50.0 \]\[ \frac{(0.0375 - 2x)^2}{(0.00125 + x)(0.00125 + x)} = 50.0 \]

Step 7: Solve for x

Solve the equation from Step 6 to find the value of \( x \). This may require solving a quadratic equation. Simplification or computational tools are useful.After solving, we find that \( x \approx 0.001035 \).

Step 8: Calculate Equilibrium Concentrations

Use \( x \) to find the concentrations at equilibrium:- \( [\mathrm{HI}] = 0.0375 - 2(0.001035) = 0.03543 \ \mathrm{mol/L} \)- \( [\mathrm{H}_2] = 0.00125 + 0.001035 = 0.002285 \ \mathrm{mol/L} \)- \( [\mathrm{I}_2] = 0.00125 + 0.001035 = 0.002285 \ \mathrm{mol/L} \)

Key Concepts

Equilibrium Constant

In chemical equilibrium, the equilibrium constant, denoted as \( K_c \), offers a quantitative measure of the proportion between products and reactants in a chemical reaction at a given temperature. Consider the example of the equilibrium \[\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g).\] At this equilibrium, the value of \( K_c = 50.0 \) at \( 745 \ \text{K} \) signifies the ratio of the concentration of \( \text{HI} \) squared to the product of the concentrations of \( \text{H}_2 \) and \( \text{I}_2 \). Understanding \( K_c \):

• A large \( K_c \) indicates a reaction close to completion, with more products than reactants.
• A small \( K_c \) shows that reactants are favored over products.
• \( K_c \) is temperature-dependent but remains constant for a given reaction at a set temperature.

To maintain equilibrium, it is crucial that the ratio defined by \( K_c \) is achieved and retained unless external conditions change.

Reaction Quotient

The reaction quotient, \( Q_c \), provides a measure of the relative quantities of products and reactants in a dynamic system at any moment in time. It is similar to the \( K_c \) expression: \[ Q_c = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} \] Using initial concentrations, you can compute \( Q_c \), which in our case is 900. Comparing \( Q_c \) with \( K_c \) allows you to determine if the system is at equilibrium or if it needs to shift to achieve equilibrium.To interpret \( Q_c \):

• If \( Q_c = K_c \), the system is at equilibrium.
• If \( Q_c > K_c \), there are more products than at equilibrium, prompting a shift towards reactants.
• If \( Q_c < K_c \), there are fewer products, requiring a shift towards products to reach equilibrium.

In our exercise, because \( Q_c > K_c \), the system has too much \( \text{HI} \) compared to \( \text{H}_2 \) and \( \text{I}_2 \), and it will shift to the left.

Le Chatelier's Principle

Le Chatelier's Principle explains how a system at equilibrium responds to disturbances or changes in conditions. When a reaction mixture is disturbed, the system adjusts to minimize that disturbance and re-establish equilibrium. This principle is crucial for predicting the direction of the reaction's shift when changes occur.In our example, since \( Q_c > K_c \), there's an excess of products. According to Le Chatelier's Principle:

• The reaction will shift to the left, favoring the formation of reactants, \( \text{H}_2 \) and \( \text{I}_2 \).
• This shift will continue until \( Q_c \) matches \( K_c \), restoring equilibrium.

This principle can also be applied when adjusting other factors like pressure, temperature, or concentration, allowing prediction of how the system will react to changes.

Concentration Calculations

Calculating concentrations in a reaction involves determining the amount of each substance per unit volume. This information is crucial in assessing the system's state and predicting its behavior.Initial concentration calculations are straightforward, using the formula:\[ C = \frac{n}{V} \]Here, \( n \) is the number of moles, and \( V \) is the volume of the container.In our problem:

• \( [\text{HI}] = \frac{0.75}{20.0} = 0.0375 \ \text{mol/L} \)
• \( [\text{H}_2] = \frac{0.025}{20.0} = 0.00125 \ \text{mol/L} \)
• \( [\text{I}_2] = 0.00125 \ \text{mol/L} \)

These initial concentrations are used to compute \( Q_c \). Then, using changes and shifts discussed earlier, we determine equilibrium concentrations. The balance use of stoichiometry and equilibrium expressions finds new concentrations satisfying \( K_c \), essential for obtaining a stable system.