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 atm, were allowed to come to equilibrium at this temperature?

Short answer

The equilibrium constant, \(K_p\), is approximately \(1.77 \times 10^{-1}\). When starting with initial partial pressures of 0.30 atm for NO and \(Br_2\), the equilibrium partial pressures are NO: \(135.6 \, \text{torr}\), \(Br_2\): \(181.8 \, \text{torr}\), and NOBr: \(92.4 \, \text{torr}\).

Step-by-step solution

Determine the initial partial pressures

Initially, we have: \( P_{NO} = 98.4 \,\text{torr} \) \( P_{Br_2} = 41.3 \,\text{torr} \) \( P_{NOBr} = 0 \)

Find the change in partial pressures as the reaction proceeds

As the reaction reaches equilibrium, let a certain amount (x) of Br_2 react: \( P_{NO} = 98.4 - 2x \,\text{torr} \) \( P_{Br_2} = 41.3 - x \,\text{torr} \) \( P_{NOBr} = 2x \,\text{torr} \) Now, we are given the total pressure at equilibrium: \( 110.5 \,\text{torr} \)

Use the total pressure to solve for the change x

At equilibrium, we can use the total pressure to find the change that has occurred: \( 110.5 = (98.4 - 2x) + (41.3 - x) + 2x \) \( x = 29.2 \,\text{torr} \)

Calculate equilibrium partial pressures

Now we can calculate the partial pressures of all the species at equilibrium: \( P_{NO} = 98.4 - 2x = 40.0 \,\text{torr} \) \( P_{Br_2} = 41.3 - x = 12.1 \,\text{torr} \) \( P_{NOBr} = 2x = 58.4 \,\text{torr} \)

Find \(K_p\)

Using the equilibrium partial pressures, we can find the value of \(K_p\): \( K_p = \frac{(P_{NOBr})^2}{(P_{NO})^2(P_{Br_2})} = \frac{(58.4)^2}{(40.0)^2(12.1)} \) \( K_p = \frac{3407.04}{19200}= 0.1773 \approx 1.77 × 10^{-1} \) #b. What would be the partial pressures of all species if NO and \(Br_2\), both at an initial partial pressure of 0.30 atm, were allowed to come to equilibrium at this temperature?# We can solve this problem by proceeding with the following steps:

Convert to the same pressure unit

Since \(K_p\) involves torr units, we first convert the given pressures in atm to torr: \(0.30 \, \text{atm} \times \frac{760 \, \text{torr}}{1 \, \text{atm}} = 228 \, \text{torr}\) Now, we have initial partial pressures: \( P_{NO} = P_{Br_2} = 228 \, \text{torr} \) \( P_{NOBr} = 0 \)

Find the change in partial pressures as the reaction proceeds

Let the change in pressures as x: \( P_{NO} = 228 - 2x \,\text{torr} \) \( P_{Br_2} = 228 - x \,\text{torr} \) \( P_{NOBr} = 2x \,\text{torr} \)

Write the expression for \(K_p\) using the expressions from Step 2 and solve for x

Using the value of \(K_p\) from part (a), we can write: \( K_p = 0.1773 = \frac{(P_{NOBr})^2}{(P_{NO})^2(P_{Br_2})} = \frac{(2x)^2}{((228-2x)^2)(228-x)} \) To solve this equation, we can employ approximation by assuming \(x\) is small compared to \(228\), allowing us to simplify the equation as: \(\frac{4x^2}{(228^2)(228)} = 0.1773\) Now we can solve for x: \( x = \sqrt{0.1773 \times (228^2)(228) \div 4} \approx 46.2 \,\text{torr} \)

Calculate equilibrium partial pressures

Using the value of x found, we can calculate the partial pressures of all species at equilibrium: \( P_{NO} = 228 - 2x = 135.6 \, \text{torr} \) \( P_{Br_2} = 228 - x = 181.8 \, \text{torr} \) \( P_{NOBr} = 2x = 92.4 \, \text{torr} \) The equilibrium partial pressures of all species for the given initial conditions are: NO: \( 135.6 \, \text{torr} \), \(Br_2\): \( 181.8 \, \text{torr} \), and NOBr: \( 92.4 \, \text{torr} \).

Key concepts

Understanding Partial Pressures

Partial pressure, a term central to gaseous reactions like the nitric oxide and bromine reaction discussed in the textbook exercise, is the pressure that a single gas in a mixture of gases would exert if it alone occupied the whole volume. When dealing with chemical reactions involving gases, understanding partial pressures is crucial because they determine how far the reaction proceeds, and are integral to calculating the equilibrium constant (Kp).

In the example provided, we determine the initial partial pressures of nitric oxide (NO) and bromine (Br₂) simply based on the given values. As the reaction proceeds towards the equilibrium, the partial pressure of each gas changes. This change is represented by 'x' - showing the consumed amount of the starting gases and the formed amount of the product NOBr. Importantly, in the context of equilibrium, the sum of the partial pressures of all components at equilibrium equals the total pressure in the system.

Achieving Chemical Equilibrium

Chemical equilibrium occurs when the rate of the forward reaction equals the rate of the reverse reaction and the concentrations of the products and reactants remain unchanged over time. It's a dynamic state, meaning the reactions are still occurring, but there's no net change in the amounts of substances involved.

In our exercise, 2 NO(g) + Br₂(g) ⇌ 2 NOBr(g), equilibrium is reached when the rate at which NO and Br₂ react to form NOBr is equal to the rate at which NOBr decomposes back into NO and Br₂. The equilibrium constant (Kp) expresses the ratio of the partial pressures of the products to those of the reactants, each raised to the power of their coefficients in the balanced equation. The value of Kp, which is a measure of the position of equilibrium, depends only on the temperature for a given reaction.

Applying Le Chatelier's Principle to Pressure Changes

Impact of Changing Conditions on Equilibrium

Le Chatelier's principle predicts how a change in conditions can shift the position of equilibrium. For gaseous reactions, changes in pressure can significantly affect the equilibrium position. If the pressure is increased, the equilibrium will shift towards the side with fewer moles of gas, as seen in our exercise.

To predict the effect of changing the initial partial pressures of NO and Br₂, Le Chatelier's principle tells us that a change to the system's conditions (like the increase in pressure due to higher starting partial pressures of the reactants) will cause the equilibrium to adjust in order to counteract the change – in this case, by shifting toward the production of more NOBr to reduce the pressure increase. By calculating the changes to the partial pressures (using the variable 'x'), we adjust for this shift and ensure the reaction maintains its new equilibrium under the modified conditions.