Question

The following equilibrium pressures at a certain temperature were observed for the reaction $$\begin{aligned} 2 \mathrm{NO}_{2}(g) & \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \\ P_{\mathrm{NO}_{2}} &=0.55 \mathrm{atm} \\\ P_{\mathrm{NO}} &=6.5 \times 10^{-5} \mathrm{atm} \\ P_{\mathrm{O}_{2}} &=4.5 \times 10^{-5} \mathrm{atm} \end{aligned}$$ Calculate the value for the equilibrium constant \(K_{\mathrm{p}}\) at this temperature.

Short answer

The value for the equilibrium constant, \(K_p\), at this temperature is approximately \(1.184 \times 10^{-9}\).

Step-by-step solution

Write the balanced chemical equation

The balanced chemical equation for the reaction is: \(2NO_2(g) \rightleftharpoons 2NO(g) + O_2(g) \) #Step 2: Write the expression for Kp#

Write the expression for Kp

The general expression for Kp for a reaction is given by the ratio of the product of the partial pressures of the products raised to their stoichiometric coefficients to the product of the partial pressures of the reactants raised to their stoichiometric coefficients. For this reaction, the expression for Kp is: \(K_p = \frac{[NO]^2 \times [O_2]}{[NO_2]^2} \) #Step 3: Substitute the given partial pressures into the Kp expression#

Substitute the given partial pressures into the Kp expression

We are given the partial pressures at equilibrium as follows: \(P_{NO_2} = 0.55 \, atm\) \(P_{NO} = 6.5 \times 10^{-5} \, atm\) \(P_{O_2} = 4.5 \times 10^{-5} \, atm\) Substitute these values into the Kp expression: \(K_p = \frac{(6.5 \times 10^{-5})^2 \times (4.5 \times 10^{-5})}{(0.55)^2} \) #Step 4: Calculate the value of Kp#

Calculate the value of Kp

Using a calculator, we can compute the value of Kp: \(K_p = \frac{(6.5 \times 10^{-5})^2 \times (4.5 \times 10^{-5})}{(0.55)^2} \approx 1.184 \times 10^{-9} \) The value for the equilibrium constant, Kp, at this temperature is approximately \(1.184 \times 10^{-9}\).

Key concepts

Equilibrium Constant (Kp)

In chemical reactions, the equilibrium constant, known as \(K_p\) for reactions involving gases, provides crucial information about the reaction's dynamics at equilibrium. The symbol \(p\) signifies that the constant is based on partial pressures instead of concentrations.
The equilibrium constant reflects the ratio between the amounts of products and reactants when the reaction has reached a state where no further change in their concentrations occurs.
To calculate \(K_p\), you set up an expression based on the balanced chemical equation. Each species is raised to the power equivalent to its stoichiometric coefficient in the equation. The result gives insight into whether products or reactants are favored at equilibrium. When \(K_p\) is large, products are favored; when small, reactants are more prevalent.

Partial Pressure

Partial pressure refers to the pressure a single gas in a mixture of gases would exert if it occupied the entire volume on its own. In equilibrium calculations, it's essential to treat each gas independently in terms of its partial pressure.
The sum of the partial pressures of all gases in a system equals the total pressure of the gas mixture. Partial pressures are vital for calculating \(K_p\). For instance, in our exercise where the given partial pressures were \(P_{NO_2} = 0.55 \, atm\), \(P_{NO} = 6.5 \times 10^{-5} \, atm\), and \(P_{O_2} = 4.5 \times 10^{-5} \, atm\), these values plug into the \(K_p\) formula to map the equilibrium state of the reaction.

Stoichiometry

Stoichiometry is the aspect of chemistry that deals with the relative quantities of reactants and products in a chemical reaction. The coefficients in a balanced chemical equation indicate the ratios in which substances react or form products.
In the context of equilibrium calculations, stoichiometry plays a vital role in forming the \(K_p\) expression. Each term in the \(K_p\) formula is raised to the power of its stoichiometric coefficient. For example, looking at our balanced equation \(2NO_2(g) \rightleftharpoons 2NO(g) + O_2(g)\), the coefficient of 2 for both NO and NO₂ means these pressures are each squared in the \(K_p\) expression, while O₂ with the coefficient of 1 is not raised to any power.

Balanced Chemical Equation

A balanced chemical equation is foundational for understanding chemical reactions. It depicts the identities and phases of reactants and products and ensures that the number of atoms for each element is the same on both sides of the equation, reflecting the law of conservation of mass.
When calculating \(K_p\), a balanced equation is essential as it helps establish the correct form of the \(K_p\) expression. The balanced equation for the exercise provided is \(2NO_2(g) \rightleftharpoons 2NO(g) + O_2(g)\).
By ensuring the equation is balanced, we translate these molar ratios into proper powers in the \(K_p\) expression and understand the proportionate amounts of gases involved, enabling precise calculations and predictions.