Problem 42 - Chemistry

Question

At \(1000 \mathrm{~K},\) a sample of pure \(\mathrm{NO}_{2}\) gas decomposes: $$2 \mathrm{NO}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g)$$ The equilibrium constant \(K_{P}\) is \(158 .\) Analysis shows that the partial pressure of \(\mathrm{O}_{2}\) is 0.25 atm at equilibrium. Calculate the pressure of \(\mathrm{NO}\) and \(\mathrm{NO}_{2}\) in the mixture.

Step-by-Step Solution

Answer

The pressure of \(NO_2\) at equilibrium is \(3 - 0.25 = 2.75\) atm, and the pressure of \(NO\) at equilibrium is \(0.5\) atm.

Step 1: Initial guess of partial pressures

Let's say that initially, before any decomposition, the pressure of \(NO_2\) is \(P\). As the reaction progresses to reach equilibrium, \(NO_2\) decomposes to form \(NO\) and \(O_2\). We know that 2 moles of \(NO_2\) give 2 moles of \(NO\) and 1 mole of \(O_2\). Let's consider that \(x\) is the amount of \(NO_2\) that dissociates at equilibrium. So, the change in pressure for \(NO_2\) is \(-x\), for \(NO\) it's \(+2x\) and for \(O_2\) it's \(+x\). Thus, at equilibrium, the pressures are: \(NO_2 (P-x)\), \(NO(2x)\), and \(O_2(x)\). But we know that the partial pressure of \(O_2\) at equilibrium is 0.25 atm, so we can say that \(x = 0.25\).

Step 2: Calculate the pressure of \(NO_2\) at equilibrium

Substitute \(x = 0.25\) into the calculated expression for \(NO_2\) equilibrium pressure. So, \(NO_2\) pressure at equilibrium is \(P - 0.25\).

Step 3: Apply the definition of the equilibrium constant \(K_P\)

The expression for the equilibrium constant \(K_P\) of the reaction is \[ K_P = \frac{{(P_{NO})^2 \cdot P_{O_2}}}{{(P_{NO_2})^2}} \] At equilibrium, we replace each \(P\) with the corresponding calculated equilibrium pressures. Therefore, \[ 158 = \frac{{(2 \cdot 0.25)^2 \cdot 0.25}}{{(P - 0.25)^2}} \]

Step 4: Solve for \(P\)

Solve the equation above for \(P\), which gives the initial pressure of \(NO_2\). After solving the equation, we find that \(P = 3\) atm.

Step 5: Calculate the pressure of \(NO\) at equilibrium

Substitute \(x = 0.25\) into the calculated expression for \(NO\) equilibrium pressure. So, \(NO\) pressure at equilibrium is \(2 \cdot 0.25 = 0.5\) atm.

Key Concepts

Partial Pressure in Reactions

Partial pressure is a way to describe the pressure exerted by a single type of gas in a mixture of gases. This concept is useful in chemical equilibrium because different gases may be present, each contributing its own share of pressure to the total. In the decomposition reaction of \({NO}_2\), where \({NO}_2\) breaks down into \({NO}\) and \({O}_2\), each gas has a distinct partial pressure at equilibrium. Consider it like a pizza sliced into parts: each gas has its slice, or pressure portion, in the overall mixture. Partial pressures are crucial because the equilibrium constant \(K_P\) for gases is calculated using these pressures. The given reaction occurs under specific conditions where \(O_2\) has a known partial pressure at equilibrium, helping us find the others. In this case, since \(O_2\)'s partial pressure is 0.25 atm, we can deduce how much \({NO}_2\) decomposed and formed other gases.

Understanding Chemical Equilibrium

Chemical equilibrium is a state where the rate of the forward reaction equals the rate of the backward reaction. This means that the amounts of reactants and products remain constant over time, although their amounts aren't necessarily equal. In the equilibrium of \(2 \mathrm{NO}_2(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{O}_2(g)\), even though decomposition happens, the concentrations of \({NO}_2\), \(NO\), and \(O_2\) become stable at their equilibrium values. When calculating pressures, the equilibrium constant \(K_P\) provides insight into the ratio of product to reactant pressures. If \(K_P\) is large, the equilibrium position favors products; if small, it favors reactants. Here, \(K_P\) at 158 suggests that products are heavily favored at 1000 K.

Gas Reactions and Equilibrium Constants

Gas reactions, like the decomposition of nitrogen dioxide \(NO_2\), often involve changes in pressure. These reactions can shift until a balance (equilibrium) is reached. The equilibrium constant \(K_P\) for such gas reactions provides a quantitative measure of this balance. It is derived from the partial pressures of the gases involved, offering a snapshot of the reaction's tendency to form products or maintain reactants. Gas reactions are dynamic, yet at equilibrium, the concentrations of gases remain fixed due to the equal rates of forward and reverse reactions. This constancy forms the foundation for calculating equilibrium constants:

• The equation takes the partial pressures and raises them to the power of their coefficients in the balanced equation.
• Substituting known pressures in \(2 \mathrm{NO}_2(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{O}_2(g)\), we determine the pressures at equilibrium.

The Decomposition of NO2

Decomposition reactions like that of \(NO_2\) involve breaking down a compound into simpler substances. Here, gaseous \(NO_2\) decomposes into \(NO\) and \(O_2\), a common process studied for its implications in atmospheric chemistry. This specific decomposition is crucial as it represents real-world scenarios where \(NO_2\), an air pollutant, transforms into other gases. In a closed system, at a high temperature of 1000 K, \(NO_2\) doesn't fully convert but reaches a stable equilibrium with its products.

• This balance depends heavily on factors like pressure and temperature.
• The initial pressure \(P\) influences how much \(NO_2\) breaks down.

Understanding this process and computing the equilibrium pressures of resulting gases helps grasp how environmental factors can influence pollution levels and aids in designing reactions for industrial applications.