Problem 22 - Chemistry The Science in Context

Question

At \(1045 \mathrm{K}\) the partial pressures of an equilibrium mixture of \(\mathrm{H}_{2} \mathrm{O}, \mathrm{H}_{2},\) and \(\mathrm{O}_{2}\) are \(0.040,0.0045,\) and \(0.0030 \mathrm{atm}\) respectively. Calculate the value of the equilibrium constant \(K_{\mathrm{p}}\) at \(1045 \mathrm{K}\) $$ 2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) $$

Step-by-Step Solution

Answer

Question: Calculate the equilibrium constant, Kp, at 1045 K for the given reaction, using the partial pressures of H2O: 0.040 atm, H2: 0.0045 atm, and O2: 0.0030 atm. Answer: The equilibrium constant, Kp, at 1045 K for the given reaction is 38.0.

Step 1: Write the expression for Kp

Since we have to calculate the equilibrium constant for the given reaction, we first write the expression for Kp using the partial pressures of the reactants and products: $$ K_\mathrm{p} = \frac{[\mathrm{H}_{2}]^2[\mathrm{O}_{2}]}{[\mathrm{H}_{2}\mathrm{O}]^2} $$

Step 2: Substitute the given partial pressures

Next, substitute the given values for the partial pressures of H2O, H2, and O2 in the Kp expression. The values are: - H2O: 0.040 atm - H2: 0.0045 atm - O2: 0.0030 atm $$ K_\mathrm{p} = \frac{[(0.0045\,\mathrm{atm})^2(0.0030\,\mathrm{atm})]}{(0.040\,\mathrm{atm})^2} $$

Step 3: Solve for Kp

Now, we need to calculate the value of Kp by solving the equation: $$ K_\mathrm{p} = \frac{[(0.0045^2)(0.0030)]}{0.040^2} $$ $$ K_\mathrm{p} = \frac{0.00006075}{0.0016} $$ $$ K_\mathrm{p} = 38.0 $$ Therefore, the value of the equilibrium constant Kp at 1045 K is 38.0.

Key Concepts

Partial Pressures in Chemical Equilibrium

Understanding partial pressures is crucial for calculating equilibrium constants in reactions that involve gases. In a mixture of gases, each type of gas exerts pressure as if it were alone in the container. This pressure is known as its partial pressure. The total pressure of the mixture is simply the sum of all the individual partial pressures of the gases present. In the exercise, the partial pressures were provided for each component of the reaction

• H2O: 0.040 atm
• H2: 0.0045 atm
• O2: 0.0030 atm

These pressures are used to calculate the equilibrium constant, illustrating how gases behave in a closed system.
Understanding how to manage and calculate these partial pressures is a key skill for understanding chemical reactions involving gases.

Reaction Calculations Using Partial Pressures

Reaction calculations in chemistry often involve using the partial pressures of reactants and products to determine the equilibrium constant, denoted as \(K_p\). This constant provides insight into how far a reaction proceeds before reaching equilibrium. For a reaction \(2 \mathrm{H}_2\mathrm{O}(g) \rightleftharpoons 2 \mathrm{H}_2(g) + \mathrm{O}_2(g)\), the equilibrium constant \(K_p\) is calculated by inserting the partial pressures of the gases into the expression:
\[K_\mathrm{p} = \frac{[\mathrm{H}_{2}]^2\,[\mathrm{O}_{2}]}{[\mathrm{H}_2 \mathrm{O}]^2}\]Reaction calculations help in understanding the proportions of reactants and products at equilibrium. By substituting the given partial pressures into this formula, one can determine \(K_p\), which in this case was found to be 38.0.
This value indicates the extent to which the reaction shifts towards products under the given conditions.

Understanding Chemical Equilibrium

Chemical equilibrium is a state where the concentrations or pressures of reactants and products remain constant over time. This occurs because the rate of the forward reaction equals the rate of the reverse reaction. In the example provided, the equilibrium constant \(K_p\) is used to quantify this state.
If \(K_p\) is much greater than 1, it suggests that the reaction heavily favors the formation of products as seen in our exercise, where \(K_p = 38.0\). Conversely, a \(K_p\) less than 1 would indicate a reaction leaning towards the reactants.
Understanding chemical equilibrium and the use of \(K_p\) helps predict the outcome of chemical reactions and thus is an invaluable concept in many scientific fields. Equilibrium calculations provide the framework to predict how changes in conditions like temperature and pressure affect the direction and extent of the reaction.