Problem 21 - Chemistry The Science in Context

Question

At \(1200 \mathrm{K}\) the partial pressures of an equilibrium mixture of \(\mathrm{H}_{2} \mathrm{S}, \mathrm{H}_{2},\) and \(\mathrm{S}\) are \(0.020,0.045,\) and 0.030 atm, respectively. Calculate the value of the following equilibrium constant at \(1200 \mathrm{K}\) $$ \mathrm{H}_{2} \mathrm{S}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{S}(g) \quad K_{\mathrm{p}}=? $$

Step-by-Step Solution

Answer

Question: Calculate the value of the equilibrium constant, Kp, for the reaction H2S(g) ⇌ H2(g) + S(g) at 1200K, given the partial pressures of H2S, H2, and S are 0.020 atm, 0.045 atm, and 0.030 atm, respectively. Answer: The value of the equilibrium constant, Kp, for the reaction H2S(g) ⇌ H2(g) + S(g) at 1200K is 0.0675.

Step 1: Write the balanced chemical equation

The balanced equation for the given reaction is: H2S(g) ⇌ H2(g) + S(g)

Step 2: Write the expression for Kp

The expression for Kp can be determined from the stoichiometry of the reaction: Kp = (P_H2 * P_S) / P_H2S

Step 3: Plug in the given partial pressures

The partial pressures for H2S, H2, and S are given as 0.020 atm, 0.045 atm, and 0.030 atm, respectively. Substitute these values into the Kp expression: Kp = (0.045 * 0.030) / 0.020

Step 4: Calculate the value of Kp

Now, we can calculate the value of Kp by performing the arithmetic operation: Kp = (0.045 * 0.030) / 0.020 Kp = 0.00135 / 0.020 Kp = 0.0675 So, the value of the equilibrium constant, Kp, for this reaction at 1200K is 0.0675.

Key Concepts

Equilibrium Constant

The equilibrium constant, usually denoted as either \(K_c\) or \(K_p\), is a measure of the extent of a chemical reaction at equilibrium. It helps to predict the concentrations or pressures of reactants and products when a reaction has reached a state of balance. In the case of gaseous reactions, \(K_p\) is used, as it is based on the partial pressures of the components involved in the reaction. The subscript "p" signifies that it relates to pressure.

The value of the equilibrium constant provides insight into the reaction's tendency to move towards products or reactants:

• A large \(K_p\) (greater than 1) indicates that, at equilibrium, the reaction mixture contains more products than reactants.
• A small \(K_p\) (less than 1) indicates more reactants than products.
• A \(K_p\) close to 1 suggests that neither reactants nor products are heavily favored.

In the given exercise, the reaction \(\mathrm{H}_{2} \mathrm{S}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{S}(g)\) at 1200 K, showed a \(K_p\) of 0.0675, indicating a moderate tendency towards forming reactants compared to products.

Partial Pressure

Partial pressure is a crucial concept in understanding chemical equilibria for reactions involving gases. It refers to the pressure that a gas in a mixture would exert if it occupied the entire volume of the mixture by itself. For gas reactions, the partial pressure of each component helps to determine the state of equilibrium using \(K_p\).

The total pressure exerted by a gas mixture is the sum of the partial pressures of each individual gas. In our exercise, the partial pressures for \(\mathrm{H}_{2} \mathrm{S}, \mathrm{H}_{2},\) and \(\mathrm{S}\) were given as 0.020 atm, 0.045 atm, and 0.030 atm, respectively.
These values are substituted into the expression for \(K_p\) as follows:
\[ K_p = \frac{(P_{\mathrm{H}_2} \times P_{\mathrm{S}})}{P_{\mathrm{H}_2\mathrm{S}}} \]where \( P_{\mathrm{H}_2} \) and \( P_{\mathrm{S}} \) are the partial pressures of \(\mathrm{H}_2\) and \(\mathrm{S}\), respectively, and \( P_{\mathrm{H}_2\mathrm{S}} \) is the partial pressure of \(\mathrm{H}_2\mathrm{S}\). Understanding how partial pressures are utilized can aid in calculating \( K_p\) and understanding equilibrium behavior.

Stoichiometry

Stoichiometry is the aspect of chemistry that deals with the quantitative relationships between the reactants and products in a chemical reaction. It is essential for writing balanced chemical equations and for deducing the expressions for equilibrium constants like \(K_p\).

In the exercise example, the balanced chemical equation is \(\mathrm{H}_{2} \mathrm{S}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{S}(g)\). The stoichiometry, using coefficients from the balanced equation, indicates how the moles and subsequently the partial pressures of the substances relate to each other at equilibrium.
For a reaction in equilibrium:

• Stoichiometry ensures the conservation of mass and charge; moles of reactants used up must equal moles of products formed.
• In the expression \(K_p = \frac{(P_{\mathrm{H}_2} \times P_{\mathrm{S}})}{P_{\mathrm{H}_2\mathrm{S}}}\), the stoichiometric coefficients from the balanced equation directly affect how the equilibrium constant is structured.

Understanding stoichiometry allows for precise manipulation and prediction of reaction behaviours and balances the scales of any given chemical change.