Problem 53 - Chemistry The Central Science In Si Units

Question

At \(900^{\circ} \mathrm{C}, K_{p}=51.2\) for the equilibrium $$ 2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) $$ If the pressure of \(\mathrm{NO}(g)\) is half the pressure of \(\mathrm{NOBr}(g)\), what is the equilibrium pressure of \(\mathrm{Br}_{2}(g)\) ?

Step-by-Step Solution

Answer

The equilibrium pressure of \(\text{Br}_2\) is \(204.8 \text{ atm}\).

Step 1: Write the Expression for Kp

The equilibrium expression for the given reaction is written using partial pressures:\[ K_{p} = \frac{{(P_{\text{NO}})^2 \cdot P_{\text{Br}_2}}}{{(P_{\text{NOBr}})^2}} \]where \(P_{\text{NO}}\), \(P_{\text{Br}_2}\), and \(P_{\text{NOBr}}\) are the equilibrium partial pressures of NO, \(\text{Br}_2\), and NOBr respectively. We know \( K_{p} = 51.2. \)

Step 2: Establish Relationships Between Pressures

Given that the pressure of NO is half the pressure of NOBr, we can define the partial pressures in terms of \(P_{\text{NOBr}}\): \[ P_{\text{NO}} = \frac{1}{2} P_{\text{NOBr}} \]Let \(P_{\text{NOBr}}= x\), so \(P_{\text{NO}} = \frac{x}{2}\).

Step 3: Substitute into Kp Expression

Substitute \(P_{\text{NO}} = \frac{x}{2}\) and \(P_{\text{NOBr}}= x\) into the expression for \(K_p\):\[ 51.2 = \frac{{\left( \frac{x}{2} \right)^2 \cdot P_{\text{Br}_2}}}{{x^2}} \]

Step 4: Simplify the Expression

Simplify the equation:\[ 51.2 = \frac{\left( \frac{x^2}{4} \right) P_{\text{Br}_2}}{x^2} \]\[ 51.2 = \frac{P_{\text{Br}_2}}{4} \]

Step 5: Solve for Equilibrium Pressure of Br2

Multiply both sides by 4 to isolate \(P_{\text{Br}_2}\):\[ 51.2 \times 4 = P_{\text{Br}_2} \]\[ P_{\text{Br}_2} = 204.8 \]

Step 6: Final Answer Calculation

The equilibrium pressure of \(\text{Br}_2\) is determined to be \(204.8 \text{ atm}\).

Key Concepts

Kp expression

The equilibrium constant, represented as \(K_p\), is a crucial concept in understanding chemical reactions at equilibrium. It is specifically used for gaseous reactions, where it is linked to the partial pressures of the reactants and products involved.Kp provides insight into the ratio of the products to reactants under equilibrium conditions.

• The expression for \(K_p\) is derived from the balanced chemical equation of the reaction.


• Consider the reaction \(2 \text{NOBr}(g) \rightleftharpoons 2 \text{NO}(g) + \text{Br}_2(g)\): here, \(K_p\) is written as: \[K_{p} = \frac{(P_{\text{NO}})^2 \cdot P_{\text{Br}_2}}{(P_{\text{NOBr}})^2}\]


• The partial pressure of each gas is raised to the power matching its coefficient in the balanced equation.

To solve problems using the \(K_p\) expression, substitute the known values of partial pressures and calculate accordingly to find unknown variables.

partial pressure

In the context of gaseous equilibria, understanding partial pressures is vital. Partial pressure refers to the pressure that a single gas in a mixture would exert if it occupied the entire volume by itself.

• Partial pressure contributes to the total pressure of the gas mixture.


• For instance, the problem establishes a relationship between the partial pressure of \(\text{NO}\) and \(\text{NOBr}\) by stating \(P_{\text{NO}} = \frac{1}{2}P_{\text{NOBr}}\).


• Utilizing such relationships allows one to express all partial pressures with respect to a single variable, simplifying the solution of equilibrium problems.

Recognizing these relationships is key to accurately solving for unknown pressures in gas equilibria.

chemical equilibrium

Chemical equilibrium occurs in reversible reactions when the rate of the forward reaction equals the rate of the reverse reaction, resulting in no net change in the concentration of reactants and products over time.

• In a state of equilibrium, concentrations of reactants and products remain constant, not necessarily equal.


• The equilibrium state is described by the equilibrium constant \(K\), and for gaseous systems, this is expressed as \(K_p\).


• Reaching equilibrium creates a balance between opposing processes, allowing predictions of how system changes (like pressure or concentration variations) will affect overall balance.

Thus, understanding chemical equilibrium helps in determining how different conditions will influence reaction extents and product ratios in chemical processes.