Problem 58 - Chemistry and Chemical Reactivity

Question

Nitrosy bromide, NOBr, dissociates readily at room temperature. $$ \operatorname{NOBr}(\mathrm{g}) \rightleftharpoons \mathrm{NO}(\mathrm{g})+1 / 2 \mathrm{Br}_{2}(\mathrm{g}) $$ Some NOBr is placed in a flask at \(25^{\circ} \mathrm{C}\) and allowed to dissociate. The total pressure at equilibrium is \(190 \mathrm{mm}\) Hg and the compound is found to be \(34 \%\) dissociated. What is the value of \(K_{\mathrm{p}} ?\)

Step-by-Step Solution

Answer

The value of \(K_p\) is calculated based on the partial pressures of the gases at equilibrium.

Step 1: Understand the Dissociation

The initial reaction involves the dissociation of NOBr into NO and half a mole of Br₂. The equation is:\[\text{NOBr(g)} \rightleftharpoons \text{NO(g)} + \frac{1}{2} \text{Br}_2(g)\]This means that for each mole of NOBr that dissociates, we get 1 mole of NO and 0.5 moles of Br₂.

Step 2: Define Initial and Change Concentrations

Assume that initially, we have 1 mole of NOBr. At equilibrium, 34% of it dissociates, meaning 0.34 moles of NOBr dissociate: - Initial NOBr = 1 mole - Dissociated NOBr = 0.34 mole So, remaining NOBr = 0.66 mole.

Step 3: Calculate Equilibrium Moles of Products

The dissociation of 0.34 moles of NOBr produces:- 0.34 moles of NO- \(0.34 \times \frac{1}{2} = 0.17\) moles of Br₂.

Step 4: Calculate Equilibrium Pressures

The total pressure at equilibrium is 190 mmHg. Since the reaction produces additional moles of gas from the dissociation, determine the partial pressures:- Total moles at equilibrium = 0.66 + 0.34 + 0.17 = 1.17- Fraction of total pressure for each component: * \[ P_{\text{NOBr}} = \frac{0.66}{1.17} \times 190 \text{ mmHg} \] * \[ P_{\text{NO}} = \frac{0.34}{1.17} \times 190 \text{ mmHg} \] * \[ P_{\text{Br}_2} = \frac{0.17}{1.17} \times 190 \text{ mmHg} \]

Step 5: Express Equilibrium Constant Kp

The expression for the equilibrium constant in terms of pressure \(K_p\) is given by:\[ K_p = \frac{P_{\text{NO}} \cdot \sqrt{P_{\text{Br}_2}}}{P_{\text{NOBr}}} \]Substitute the calculated partial pressures into the equation to find \(K_p\).

Key Concepts

Dissociation Reaction

Dissociation reactions occur when a compound breaks down into two or more simpler substances. In the case of nitrosyl bromide (NOBr), it dissociates into nitrogen monoxide (NO) and bromine (Br₂) gas. This process can be represented by: \[\operatorname{NOBr} (\text{g}) \rightleftharpoons \operatorname{NO} (\text{g}) + \frac{1}{2} \operatorname{Br}_2 (\text{g})\] Understanding this reaction is crucial as it directly affects how we calculate the equilibrium state of the system. The stoichiometry of the reaction indicates that one mole of NOBr will yield one mole of NO and half a mole of Br₂. This stoichiometric relationship is essential for determining how much of each substance is present at equilibrium.

Partial Pressure

In a gaseous system, the pressure exerted by each individual gas in a mixture is called partial pressure. It is a vital concept in reactions involving gases, like the dissociation of NOBr into NO and Br₂. The total pressure at equilibrium is the sum of the partial pressures of all gases present.
To find the partial pressure of each component in the reaction, use their mole fractions relative to the total moles in the mixture. For example, if 34% of NOBr dissociates, calculate the pressure each gas contributes using its share of the total moles. This step is crucial in determining the equilibrium constant, as it relies on knowing these partial pressures.

Equilibrium Constant

The equilibrium constant \( K_p \) for a reaction in terms of pressure is a measure of the ratio of the products' partial pressures to the reactants' partial pressures at equilibrium. It’s a fundamental value that indicates the extent of a reaction under given conditions.
For the dissociation of NOBr, the expression for \( K_p \) is: \[ K_p = \frac{P_{\text{NO}} \cdot \sqrt{P_{\text{Br}_2}}}{P_{\text{NOBr}}}\] By substituting the partial pressures of NO, Br₂, and NOBr into this equation, you can find \( K_p \). This calculation tells us how the reaction behaves at a particular temperature, providing insight into which direction the reaction favors, completion or reactant presence.

Gaseous Equilibrium

Gaseous equilibrium refers to the state where the rate of the forward reaction equals the rate of the reverse reaction in a closed system involving gases. At this point, the concentrations of reactants and products remain constant over time.
In the NOBr dissociation reaction, the system achieves equilibrium when the amount of NOBr decomposing equals the amount re-forming from NO and Br₂. Understanding gaseous equilibrium helps in predicting how changes in pressure, temperature, or concentration might affect the system.
For example, at equilibrium, if the total pressure is manipulated, according to Le Chatelier's Principle, the system will adjust to counterbalance the change, either by forming more products or reactants. This concept is fundamental when discussing equilibrium shifts and influences real-world applications like chemical reactors.