Question

At \(373 \mathrm{~K}, K_{p}=0.416\) for the equilibrium $$ 2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) $$ If the pressures of \(\operatorname{NOBr}(g)\) and \(\mathrm{NO}(g)\) are equal, what is the equilibrium pressure of \(\mathrm{Br}_{2}(g)\) ?

Short answer

The equilibrium pressure of Br₂(g) can be expressed in terms of the pressure of NO(g) as: $$ P_{Br_{2}}=0.416 \cdot (P_{NO})^2 $$ Once the pressure of NO(g) is known, we can find the equilibrium pressure of Br₂(g).

Step-by-step solution

Write the equilibrium expression

For the given reaction, the equilibrium expression in terms of partial pressures (Kp) can be written as: $$ K_{p} = \frac{P_{NO}^2 \cdot P_{Br2}}{P_{NOBr}^2} $$ Where \(P_{NO}\), \(P_{Br2}\), and \(P_{NOBr}\) are the partial pressures of NO, Br2, and NOBr respectively at equilibrium.

Substitute the given value of Kp and equal partial pressures

We are given \(K_p = 0.416\) at 373K and the pressures of NOBr(g) and NO(g) are equal. Let the equal partial pressures be denoted as \(P\). The equilibrium expression becomes: $$ 0.416 = \frac{P^2 \cdot P_{Br2}}{P^2} $$

Solve for the equilibrium pressure of Br₂(g)

Let's solve the equation for \(P_{Br2}\): $$ P_{Br2} = 0.416 \cdot P^2 $$ Since the pressures of NOBr and NO are equal, we can substitute \(P\) from the given equation in terms of \(P_{NO}\) and \(P_{NOBr}\). We know that \(P_{NO} = P_{NOBr}\): $$ P_{Br2} = 0.416 \cdot (P_{NO})^2 $$ We cannot find the exact value of \(P_{Br2}\) since we do not have the value for \(P_{NO}\). However, we have expressed the equilibrium pressure of Br₂(g) in terms of the pressure of NO(g). Once the pressure of NO(g) is known, we can easily find the pressure of Br₂(g).