Problem 80 - Chemistry The Science in Context

Question

Passing steam over hot carbon produces a mixture of carbon monoxide and hydrogen known as water gas: $$ \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{C}(s) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2}(g) $$ The value of \(K_{\mathrm{c}}\) for the reaction at \(1000^{\circ} \mathrm{C}\) is \(3.0 \times 10^{-2}\) a. Calculate the equilibrium partial pressures of the products and reactants at \(1000^{\circ} \mathrm{C}\) if \(P_{\mathrm{H}_{2} \mathrm{O}}=0.442\) atm and \(P_{\mathrm{CO}}=5.0\) atm at the start of the reaction. Assume that the carbon is in excess. b. Determine the equilibrium partial pressures of the reactants and products after sufficient \(\mathrm{CO}\) and \(\mathrm{H}_{2}\) are added to the equilibrium mixture in part (a) to initially increase the partial pressures of both gases by 0.075 atm.

Step-by-Step Solution

Answer

Based on the given value of Kc and the step-by-step solution, the equilibrium partial pressures at 1000°C are as follows: For part a: \(P_{H_2O} \approx 0.428 \ \mathrm{atm}\) \(P_{CO} \approx 5.014 \ \mathrm{atm}\) \(P_{H_2} \approx 0.0135 \ \mathrm{atm}\) For part b, after the increase in partial pressures of CO and H₂ by 0.075 atm: \(P_{H_2O} \approx 0.4259 \ \mathrm{atm}\) \(P_{CO} \approx 5.0869 \ \mathrm{atm}\) \(P_{H_2} \approx 0.0906 \ \mathrm{atm}\)

Step 1: Write the expression for Kc

For the reaction given, the equilibrium constant \(K_{\mathrm{c}}\) can be expressed as: $$ K_\mathrm{c} = \frac{[\mathrm{CO}] \cdot [\mathrm{H}_2]}{[\mathrm{H}_2\mathrm{O}]} $$

Step 2: Create the ICE table for part a

ICE table for the given reaction is: Initial: \( [\mathrm{H_2O}] = 0.442 \mathrm{atm}\) ; \([\mathrm{CO}] = 5.0 \mathrm{atm}\) ; \([\mathrm{H_2}] = 0\) Change: \( -x\) ; \(x\) ; \(x\) Equilibrium: \( (0.442-x)\) ; \((5 + x)\) ; \(x\)

Step 3: Substitute the values into the Kc expression

Now we substitute the equilibrium values into the Kc expression and solve for x: $$ 3.0 \times 10^{-2} = \frac{(5 + x) \cdot x}{(0.442 - x)} $$

Step 4: Solve for x

Solving this equation for x, we get: \(x \approx 0.0135\)

Step 5: Calculate equilibrium partial pressures for part a

Now we can calculate the equilibrium partial pressures: \(P_{H_2O} = 0.442 - 0.0135 \approx 0.428 \ \mathrm{atm}\) \(P_{CO} = 5.0 + 0.0135 \approx 5.014 \ \mathrm{atm}\) \(P_{H_2} = 0.0135 \ \mathrm{atm}\)

Step 6: Create the ICE table for part b

For part b, we add 0.075 atm to the initial partial pressures of \(\mathrm{CO}\) and \(\mathrm{H}_{2}\): Initial: \( [\mathrm{H_2O}] = 0.428 \mathrm{atm}\) ; \([\mathrm{CO}] = 5.089 \mathrm{atm}\) ; \([\mathrm{H_2}] = 0.0885 \mathrm{atm}\) Change: \( -y\) ; \(-y\) ; \(y\) Equilibrium: \( (0.428-y)\) ; \((5.089 - y)\) ; \((0.0885 + y)\)

Step 7: Substitute the values into the Kc expression

Now we substitute the equilibrium values into the Kc expression and solve for y: $$ 3.0 \times 10^{-2} = \frac{(5.089 - y) \cdot (0.0885 + y)}{(0.428 - y)} $$

Step 8: Solve for y

Solving this equation for y, we get: \(y \approx 0.00209\)

Step 9: Calculate equilibrium partial pressures for part b

Now we can calculate the new equilibrium partial pressures: \(P_{H_2O} = 0.428 - 0.00209 \approx 0.4259 \ \mathrm{atm}\) \(P_{CO} = 5.089 - 0.00209 \approx 5.0869 \ \mathrm{atm}\) \(P_{H_2} = 0.0885 + 0.00209 \approx 0.0906 \ \mathrm{atm}\)

Key Concepts

Equilibrium Constant

Understanding the equilibrium constant \(K_c\) is crucial in chemical equilibrium. It measures the relative concentrations of products and reactants for a reaction at equilibrium. For the reaction \(\mathrm{H_2O}(g) + \mathrm{C}(s) \rightleftharpoons \mathrm{CO}(g) + \mathrm{H_2}(g)\), the equilibrium constant \(K_c\) is given by:

• \(K_\mathrm{c} = \frac{[\mathrm{CO}] \cdot [\mathrm{H}_2]}{[\mathrm{H}_2\mathrm{O}]}\)

Here, the square brackets \([ ]\) indicate partial pressure measurements in atm. This value helps predict how a reaction proceeds.
At \(1000^{\circ} \mathrm{C}\), \(K_c\) is \(3.0 \times 10^{-2}\), indicating a small amount of products compared to reactants at equilibrium.
A higher \(K_c\) means more products are formed.

ICE Table

The ICE table is a powerful tool for calculating changes in concentration or pressure when a system reaches equilibrium.
It stands for Initial, Change, and Equilibrium:

• Initial: The starting concentrations or pressures of all species.
• Change: The change in concentrations or pressures to reach equilibrium.
• Equilibrium: The values after equilibrium is established.

For example, in part a of the exercise, the initial conditions are \(P_{\mathrm{H_2O}} = 0.442\) atm, \(P_{\mathrm{CO}} = 5.0\) atm, and \(P_{\mathrm{H_2}} = 0\) atm. Changes are represented by \(x\), and the equilibrium pressures are expressed in terms of \(x\).
This method streamlines solving equilibrium problems.

Partial Pressure Calculation

Calculating partial pressures at equilibrium requires substituting values from the ICE table into the equilibrium expression.
For part a, you solve:\[3.0 \times 10^{-2} = \frac{(5 + x) \cdot x}{(0.442 - x)} \]Solving this for \(x\), you find \(x \approx 0.0135\).
You then calculate equilibrium pressures:

• \(P_{\mathrm{H_2O}} = 0.442 - 0.0135 \approx 0.428\) atm
• \(P_{\mathrm{CO}} = 5.0 + 0.0135 \approx 5.014\) atm
• \(P_{\mathrm{H_2}} = 0.0135\) atm

In part b, changes are introduced by adding gases, and a similar process is followed. This approach reveals how pressures adjust to maintain equilibrium.

Le Chatelier's Principle

Le Chatelier's Principle helps predict how a system at equilibrium responds to changes. It states that if a system at equilibrium is disturbed, it will shift to counteract the disturbance and restore equilibrium.
In the context of the exercise, adding more \(\mathrm{CO}\) and \(\mathrm{H_2}\) increased their partial pressures by 0.075 atm.
According to Le Chatelier, the equilibrium will shift to the left to reduce the added gases.

• This results in more \(\mathrm{H_2O}\) being formed, decreasing the added \(\mathrm{CO}\) and \(\mathrm{H_2}\).

Understanding this principle aids in predicting and controlling reaction outcomes in various chemical processes.