At \(1000 \mathrm{~K}\), a sample of pure \(\mathrm{NO}_{2}\) gas decomposes as: $$ 2 \mathrm{NO}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) $$ The equilibrium constant \(K_{p}\) is \(156.25\) atm. Analysis shows that the partial pressure of \(\mathrm{O}_{2}\) is \(0.25\) atm at equilibrium. The partial pressure of \(\mathrm{NO}_{2}\) at equilibrium is : (a) \(0.01\) (b) \(0.02\) (c) \(0.04\) (d) None of these

The equilibrium constant, symbolized as K or Kp when using partial pressures, is a quantitative measure of the position of the equilibrium in a chemical reaction. It provides us with a snapshot of the relative concentrations or pressures of reactants and products at equilibrium conditions.

For the reaction at hand:
2NO₂(g) ⇌ 2NO(g) + O₂(g)
The equilibrium constant expression in terms of partial pressures, Kp, is $$K_{p} = \frac{(P_{\text{NO}})^2 \times P_{\text{O}_2}}{(P_{\text{NO}_2})^2}$$Here, the constant is a fixed value at a given temperature (1000 K in the exercise), which in this case is 156.25 atm. It indicates the extent of the reaction; a higher value suggests products are favored, while a lower value suggests reactants are favored. The magnitude of Kp is crucial for predicting the direction of the reaction under different conditions.

Partial pressure refers to the pressure that a single gas in a mixture would exert if it were alone in the entire volume of the mixture. This concept is vital when dealing with gas-phase reactions at equilibrium. For each species in a reaction, their respective partial pressure is denoted by P with a subscript of the compound's formula.

In the context of the given reaction, the partial pressure of NO₂ and its relation to NO and O₂ at equilibrium is essential to finding the unknown values. The term 'partial' signifies that it's only part of the total pressure exerted by all gases present.

The equilibrium expression is a formula that allows us to relate the concentrations or partial pressures of reactants and products at equilibrium for a particular reaction. Setting up the correct expression is crucial to calculate K or Kp.

In our solved exercise, the equilibrium expression was written as a function of the partial pressures of the gases involved, considering the stoichiometry of the balanced equation. Recognizing that each NO₂ molecule that decomposes produces both NO and O₂ simplifies calculations. The exercise successfully demonstrated how to manipulate the expression using the known value of Kp and the given partial pressures to solve for unknowns.