Problem 93 - Chemistry: An Atoms-Focused Approach

Question

NO, Pollution \(\ln\) a study of the formation of \(\mathrm{NO}_{x}\) air pollution, a chamber heated to \(2200^{\circ} \mathrm{C}\) was filled with air \(\left(0.79 \mathrm{atm} \mathrm{N}_{2}, 0.21 \mathrm{atm} \mathrm{O}_{2}\right) .\) What are the equilibrium partial pressures of \(\mathrm{N}_{2}, \mathrm{O}_{2},\) and \(\mathrm{NO}\) if \(K_{\mathrm{p}}=0.050\) for the following reaction at \(2200^{\circ} \mathrm{C} ?\) $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) $$

Step-by-Step Solution

Answer

Question: Calculate the equilibrium partial pressures of N₂, O₂, and NO, given initial partial pressures of N₂ as 0.79 atm and O₂ as 0.21 atm in a chamber at 2200°C and Kp = 0.050. Answer: The equilibrium partial pressures are approximately: N₂ = 0.755 atm O₂ = 0.175 atm NO = 0.070 atm

Step 1: List initial partial pressures and given \(K_{\mathrm{p}}\) value

Given initial partial pressures are: \(\mathrm{N}_{2} = 0.79\,\mathrm{atm}\) \(\mathrm{O}_{2} = 0.21\,\mathrm{atm}\) Initial partial pressure of \(\mathrm{NO} = 0\,\mathrm{atm}\), since it has not been formed yet. Also given, \(K_{\mathrm{p}} = 0.050\).

Step 2: Define the changes in partial pressures that occur at equilibrium

Let the change in the partial pressure for the reactants (\(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\)) be \(x\). Then at equilibrium, the change in partial pressure for the product \(\mathrm{NO}\) would be \(2x\). Therefore: At equilibrium: \(\mathrm{N}_{2} = 0.79 - x\,\mathrm{atm}\) \(\mathrm{O}_{2} = 0.21 - x\,\mathrm{atm}\) \(\mathrm{NO} = 2x\,\mathrm{atm}\)

Step 3: Write the expression for \(K_{\mathrm{p}}\) and substitute the equilibrium partial pressures

The expression for \(K_{\mathrm{p}}\) is given by: \(K_{\mathrm{p}} = \frac{[\mathrm{NO}]^2}{[\mathrm{N}_{2}][\mathrm{O}_{2}]}\) Substitute the equilibrium partial pressures into the expression: \(0.050 = \frac{(2x)^2}{(0.79-x)(0.21-x)}\)

Step 4: Solve for x

Solve the equation for \(x\): \(0.050(0.79-x)(0.21-x) = (2x)^2\) You can use either a numerical method (such as the quadratic formula or a calculator) or an algebraic method (such as factoring) to solve for \(x\). In this case, it's easier to use a numerical method: For this exercise, we will use a calculator to find that the numerical solution for \(x\) is \(x \approx 0.035\).

Step 5: Calculate the equilibrium partial pressures

With the calculated value of \(x\), we can now find the equilibrium partial pressures of \(\mathrm{N}_{2}, \mathrm{O}_{2},\) and \(\mathrm{NO}\): \(\mathrm{N}_{2} = 0.79 - x \approx 0.79 - 0.035 = 0.755\,\mathrm{atm}\) \(\mathrm{O}_{2} = 0.21 - x \approx 0.21 - 0.035 = 0.175\,\mathrm{atm}\) \(\mathrm{NO} = 2x \approx 2(0.035) = 0.070\,\mathrm{atm}\) Thus, the equilibrium partial pressures are approximately: \(\mathrm{N}_{2} = 0.755\,\mathrm{atm}\) \(\mathrm{O}_{2} = 0.175\,\mathrm{atm}\) \(\mathrm{NO} = 0.070\,\mathrm{atm}\)

Key Concepts

Partial Pressure

Partial pressure refers to the pressure that a single gas in a mixture of gases would exert if it alone occupied the entire volume. In any given container, the total pressure exerted by the gas mixture is the sum of the partial pressures of each individual gas.
For instance, in our exercise, nitrogen (\(\mathrm{N}_2\)) and oxygen (\(\mathrm{O}_2\)) are initially present with known partial pressures of 0.79 atm and 0.21 atm respectively.

• This means each gas contributes to the total pressure independently of the others.
• The initial partial pressure of \(\mathrm{NO}\) is 0 atm, as it's yet to be formed.

Partial pressures are crucial for calculating equilibrium states in chemical reactions involving gases.

Chemical Equilibrium

In chemical reactions, equilibrium is the state at which the concentrations of all reactants and products remain constant over time. This occurs because the forward and reverse reactions proceed at the same rate.
In our scenario, the reaction between \(\mathrm{N}_2\) and \(\mathrm{O}_2\) to form \(\mathrm{NO}\) achieves equilibrium in the closed system at the given temperature.

• Equilibrium does not mean that the reactants and products are equal in concentration, but that their rates of formation are balanced.
• Once equilibrium is reached, the partial pressures of each gas no longer change.

Finding these equilibrium values allows us to understand how the chemical system behaves under certain conditions.

Reaction Quotient

The reaction quotient, denoted as \(Q\), is used to determine the direction in which a reaction needs to proceed to reach equilibrium. It has the same form as the equilibrium constant expression, but it uses the initial or non-equilibrium concentrations or pressures.
For our reaction:

• The expression for \(K_{\mathrm{p}}\) (which is an equilibrium constant involving partial pressures) is \(K_{\mathrm{p}} = \frac{[\mathrm{NO}]^2}{[\mathrm{N}_2][\mathrm{O}_2]}\).
• At equilibrium, \(Q = K_{\mathrm{p}}\).

By comparing \(Q\) to \(K_{\mathrm{p}}\), you can predict whether the reaction will proceed forward or backward to reach equilibrium. In this exercise, solving for the change in partial pressures allows us to determine that equilibrium was achieved with the calculated partial pressures of \(\mathrm{N}_2\), \(\mathrm{O}_2\), and \(\mathrm{NO}\).