Question

Nitric oxide and bromine at initial partial pressures of 98.4 and 41.3 torr, respectively, were allowed to react at \(300 .\) K. At equilibrium the total pressure was 110.5 torr. The reaction is $$2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g)$$ a. Calculate the value of \(K_{\mathrm{p}}\) . b. What would be the partial pressures of all species if NO and \(\mathrm{Br}_{2},\) both at an initial partial pressure of \(0.30 \mathrm{atm},\) were allowed to come to equilibrium at this temperature?

Short answer

a. The equilibrium constant, Kp, for the given reaction is 21.0. b. The partial pressures at equilibrium for NO, Br2, and NOBr in the new scenario are 41.4 torr, 134.7 torr, and 186.6 torr, respectively.

Step-by-step solution

Calculate the initial pressure of NOBr

At the beginning of the reaction, NOBr is not present, so the initial pressure of NOBr is 0 torr.

Calculate the change in pressure

At equilibrium, the total pressure is given as 110.5 torr. Let's denote the change in pressure for the reaction as "x". We can express the change in pressure in terms of the coefficients of each component in the balanced equation: Initial pressures: NO = 98.4, Br2 = 41.3, and NOBr = 0. Change in pressures: ∆NO = -2x, ∆Br2 = -x, and ∆NOBr = 2x.

Calculate the equilibrium pressures

We calculate the equilibrium pressures of all the components by adding the initial pressures and the change in pressures: Equilibrium pressures: NO = 98.4 - 2x, Br2 = 41.3 - x, and NOBr = 0 + 2x. Since the total pressure at equilibrium is 110.5 torr: 98.4 - 2x + 41.3 - x + 2x = 110.5

Solve for x

Solve for x to get the change in pressure: 139.7 - x = 110.5 x = 29.2 torr

Calculate equilibrium pressures

Now we can find the equilibrium pressures of the components by substituting the value of x: NO = 98.4 - 2(29.2) = 40.0 torr Br2 = 41.3 - 29.2 = 12.1 torr NOBr = 2(29.2) = 58.4 torr

Find Kp

Now we can use the equilibrium constant formula to find Kp: Kp = [(NOBr)^2] / [(NO)^2 * (Br2)] Kp = [(58.4)^2] / [(40.0)^2 * (12.1)] = 21.0 So, Kp = 21.0. #b. Calculating partial pressures at equilibrium#

Express initial pressures

Given the initial pressures of NO and Br2 as 0.30 atm, we can express it in terms of torr: 0.30 atm × (760 torr/1 atm) = 228 torr

Use Kp to calculate x

With the new initial pressures, Kp is given by: Kp = [2x]^2 / [(228 - 2x)^2 * (228 - x)] 21.0 = (4x^2) / (4x^3 - 912x^2 + 51984x)

Solve for x

We can denote the initial number of moles as NO = Br2 = n, then we have: 21.0 = (4(29.2)^2) / (4(29.2)^3 - (912 × 29.2^2) × n) Solve for x = 93.3 torr

Calculate equilibrium pressures

Now we can find the equilibrium pressures of the components by substituting the value of x in the expressions we found earlier: NO = 228 - 2(93.3) = 41.4 torr Br2 = 228 - 93.3 = 134.7 torr NOBr = 2(93.3) = 186.6 torr So, the partial pressures at equilibrium are NO = 41.4 torr, Br2 = 134.7 torr, and NOBr = 186.6 torr.

Key concepts

Partial Pressure

Partial pressure refers to the pressure that a specific gas in a mixture would exert if it occupied the entire volume by itself. It's a vital concept for understanding how gases behave in reactions. To calculate it, you can use Dalton's Law of Partial Pressures, which states that the total pressure of a mixture of gases is equal to the sum of the partial pressures of each gas. For example, in the given reaction, we first have initial partial pressures for NO and Br\(_2\).

• Initial Partial Pressures: These are the pressures of each gas before the reaction begins. In our example, NO starts at 98.4 torr, while Br\(_2\) begins at 41.3 torr.
• Change in Partial Pressures: As the reaction proceeds, these pressures change according to the stoichiometry of the reaction. For instance, NO and Br\(_2\) decrease, while NOBr increases.
• Equilibrium Partial Pressures: These are the pressures of each gas when the reaction has reached equilibrium, meaning the rate of the forward reaction equals the rate of the reverse reaction.

Grasping partial pressures helps in understanding how the reactant and product gases evolve over the course of the reaction.

Reaction Equilibrium

Reaction equilibrium occurs when a chemical reaction and its reverse are happening at the same rate. At this point, the concentrations (or pressures, in the case of gases) of the reactants and products remain constant. The equilibrium constant, \( K_p \), helps us understand the proportion of products to reactants at equilibrium using the formula:
\[K_{p} = \frac{{[\text{NOBr}]^2}}{{[\text{NO}]^2 \times [\text{Br}_2]}}\]
This equation uses the partial pressures of the gases involved in the reaction. A high \( K_p \) value indicates a reaction favoring products, while a low value favors reactants.

• Achieving Equilibrium: In our example, the pressures change until the system stabilizes at equilibrium.
• Calculating \( K_p \): Using the equilibrium pressures in the formula allows us to find the equilibrium constant, providing insights into the reaction's tendencies at the given conditions.
• Understanding Conditions: Equilibrium shifts with changes in temperature, pressure, or concentration, which can be predicted using Le Chatelier's principle.

Understanding reaction equilibrium enables you to predict how different conditions affect the balance between reactants and products.

Gas Laws

Gas laws describe the behavior of gases, including relationships between volume, temperature, pressure, and amount. Understanding these laws is essential for predicting how gases will interact under different circumstances, such as in chemical reactions.
Ideal Gas Law: This can be expressed as \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is moles, \( R \) is the gas constant, and \( T \) is temperature. It's useful for predicting how gases will behave in a container.
Dalton’s Law: Focuses on partial pressures, stating that the total pressure of a gaseous mixture is equal to the sum of the individual gas partial pressures.

• Conversion Between Units: Often, pressures may need to be converted between measurement units such as atm and torr.
• Applications to Equilibria: In our reaction, understanding how gas laws apply can help predict how changing one variable, like temperature or pressure, could affect the whole system.
• Practical Use: These laws can predict real-world reactions, make calculations more accessible, and provide deeper understanding of gas behaviors in closed systems.

Grasping gas laws allows students to connect theoretical concepts with experimental situations, making the behavior of gases in reactions easier to predict and understand.