Problem 89 - Chemistry The Molecular Science

Question

The equilibrium constant \(K_{\mathrm{c}}\) is \(1.6 \times 10^{5}\) at \(1297 \mathrm{~K}\) and \(3.5 \times 10^{4}\) at \(1495 \mathrm{~K}\) for the reaction $$ \mathrm{H}_{2}(\mathrm{~g})+\mathrm{Br}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HBr}(\mathrm{g}) $$ (a) Is \(\Delta_{r} H^{\circ}\) for this reaction positive or negative? (b) Calculate \(K_{\mathrm{c}}\) at \(1297 \mathrm{~K}\) for the reaction $$ \frac{1}{2} \mathrm{H}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{Br}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{HBr}(\mathrm{g}) $$ (c) Pure HBr is placed into an evacuated container of constant volume. The container is sealed and heated to \(1297 \mathrm{~K} .\) Calculate the percentage of HBr that is decomposed to \(\mathrm{H}_{2}\) and \(\mathrm{Br}_{2}\) at equilibrium.

Step-by-Step Solution

Answer

(a) \(\Delta_{r} H^{\circ}\) is negative. (b) \(K_{c} = 400\). (c) \(0.79\%\) of HBr decomposes.

Step 1: Determine the sign of \(\Delta_{r} H^{\circ}\)

The equilibrium constant \(K_{c}\) decreases as temperature increases (from \(1.6 \times 10^5\) at \(1297 \mathrm{~K}\) to \(3.5 \times 10^4\) at \(1495 \mathrm{~K}\)). According to the van 't Hoff equation, if the equilibrium constant decreases with increasing temperature, the reaction is exothermic. Hence, \(\Delta_{r} H^{\circ}\) is negative.

Step 2: Calculate \(K_{c}\) at 1297 K for the half-reaction

For the reaction \(\mathrm{H}_{2} + \mathrm{Br}_{2} \rightleftharpoons 2 \mathrm{HBr}\), \(K_{c} = 1.6 \times 10^5\) at 1297 K. For the reaction \(\frac{1}{2} \mathrm{H}_{2} + \frac{1}{2} \mathrm{Br}_{2} \rightleftharpoons \mathrm{HBr}\), the equilibrium constant \(K'_{c}\) is the square root of the original \(K_{c}\) because the stoichiometry is halved.\[ K'_{c} = \sqrt{1.6 \times 10^5} \approx 400 \]

Step 3: Calculate the percentage of HBr decomposed

Starting with pure HBr, at equilibrium, let \(x\) be the amount of HBr that decomposes to form \(\mathrm{H}_{2}\) and \(\mathrm{Br}_{2}\). The equilibrium expression for the reaction \(\mathrm{H}_{2} + \mathrm{Br}_{2} \rightleftharpoons 2 \mathrm{HBr}\) is: \[ K_{c} = \frac{[\mathrm{HBr}]^2}{[\mathrm{H}_{2}][\mathrm{Br}_{2}]} \]Let initial concentration of HBr be \([\mathrm{HBr}]_0 = 1\) (assuming a basis). Therefore, at equilibrium, \([\mathrm{HBr}] = 1 - x\), \([\mathrm{H_{2}}] = x\), and \([\mathrm{Br_{2}}] = x\). Substituting these values:\[ 1.6 \times 10^5 = \frac{(1 - x)^2}{x^2} \]Solving for \(x\) yields:\[ (1 - x)^2 = 1.6 \times 10^5 \cdot x^2 \]The quadratic is solved to find \(x \approx 0.0079\), meaning \((1-0.0079) = 0.9921\) fraction of HBr remains, thus \(\approx 0.79\% \) of HBr decomposes.

Key Concepts

Equilibrium Constant

The equilibrium constant, often represented as \(K_{c}\) or \(K_{p}\) for pressure, is a number that quantifies the ratio of concentrations of products to reactants at equilibrium for a reversible chemical reaction. It's crucial because it informs us about the position of equilibrium.
A large equilibrium constant means the reaction heavily favors the formation of products, whereas a small one favors the reactants.
For example, in the reaction \( \text{H}_2 + \text{Br}_2 \rightleftharpoons 2 \text{HBr} \), a \(K_{c}\) of \(1.6 \times 10^5\) indicates a strong preference for products at \(1297 \text{ K}\).
Understanding \(K_{c}\) is essential for predicting how a reaction will respond to changes in conditions, such as temperature and pressure. As in this scenario, observing how the equilibrium constant changes with temperature can also suggest the nature of the reaction, whether exothermic or endothermic.

Van 't Hoff Equation

The Van 't Hoff equation relates the change in the equilibrium constant \(K\) of a reaction to the change in temperature, providing insights into the thermodynamics of the process. This equation is given by:\[\ln \left(\frac{K_2}{K_1}\right) = \frac{-\Delta_{r}H^{\circ}}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)\]where \(\Delta_{r}H^{\circ}\) represents the reaction's standard enthalpy change, \(R\) is the gas constant, and \(T_1\) and \(T_2\) are different temperatures.
In the problem at hand, the equilibrium constant decreases as temperature increases, suggesting that the reaction releases heat, making it exothermic. This behavior happens because an exothermic reaction will shift towards the reactants as temperature rises, reducing \(K\).

Exothermic Reaction

An exothermic reaction is one that releases heat to its surroundings, usually resulting in a temperature increase in the environment. This means that the enthalpy change \(\Delta_{r}H^{\circ}\) is negative. Typical examples include combustion reactions, such as burning wood or gasoline.
In the context of the given problem, because the equilibrium constant \(K_c\) decreases with a rise in temperature, it suggests that the reaction \(\text{H}_2 + \text{Br}_2 \rightleftharpoons 2 \text{HBr} \) is exothermic. This aligns with Le Chatelier's principle, which states that a system will shift to counteract a change (in this case, an increase in temperature) by favoring the formation of reactants, hence why \(K_c\) is smaller at higher temperatures. Understanding this concept helps predict and control reactions in industrial processes, where managing heat is crucial for safety and efficiency.