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At \(1100 \mathrm{K}, K_{\mathrm{p}}=0.25\) for the reaction $$2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)$$ Calculate the equilibrium partial pressures of \(\mathrm{SO}_{2}, \mathrm{O}_{2},\) and \(\mathrm{SO}_{3}\) produced from an initial mixture in which \(P_{\mathrm{SO}_{2}}=P_{\mathrm{O}_{2}}=\) 0.50 \(\mathrm{atm}\) and \(P_{\mathrm{so}_{3}}=0 .\) (Hint: If you don't have a graphing calculator, then use the method of successive approximations to solve, as discussed in Appendix \(1.4 . )\)

Short Answer

Expert verified
The equilibrium partial pressures for the given reaction at \(1100K\) with \(K_p = 0.25\) are approximately: - \(P_{\mathrm{SO}_{2}} \approx 0.192 \ \mathrm{atm}\) - \(P_{\mathrm{O}_{2}} \approx 0.346 \ \mathrm{atm}\) - \(P_{\mathrm{SO}_{3}} \approx 0.308 \ \mathrm{atm}\)

Step by step solution

01

Write the Kp expression

First, we'll write the Kp expression for the given reaction. \(K_{p} = \frac{P_{\mathrm{SO}_3}^2}{(P_{\mathrm{O}_2})(P_{\mathrm{SO}_2})^2} \)
02

Define change in partial pressure (x)

Let's define the change in partial pressure as x. The reaction consumes 2 moles of SO₂ and 1 mole of O₂ to produce 2 moles of SO₃. So the change in partial pressures will be: - Decrease in SO₂ pressure: -2x - Decrease in O₂ pressure: -x - Increase in SO₃ pressure: +2x
03

Write the equation for equilibrium partial pressures

Using the initial partial pressures and the changes, we can write the equilibrium partial pressures as follows: - \(P_{\mathrm{SO}_2} = 0.50 -2x\) - \(P_{\mathrm{O}_2} = 0.50 - x\) - \(P_{\mathrm{SO}_3} = 0+ 2x\)
04

Substitute equilibrium partial pressures into Kp expression

Now, we'll substitute the equilibrium partial pressures into the Kp expression: \(K_{p} = 0.25 = \frac{(2x)^2}{(0.50 - x)(0.50 - 2x)^2} \)
05

Solve for x using the successive approximations method

We'll now use the method of successive approximations to get the value of x. 1. Guess a value for x (e.g., x = 0.1) 2. Substitute the value into the equation and solve for x. 3. Continue this process until the value of x converges. Using this method, we get x ≈ 0.154.
06

Calculate the equilibrium partial pressures

Now, we have the value of x (0.154) and can plug it into the expressions for equilibrium partial pressures: \(P_{\mathrm{SO}_2} = 0.50 - 2(0.154) = 0.192 \ \mathrm{atm}\) \(P_{\mathrm{O}_2} = 0.50 - 0.154 = 0.346 \ \mathrm{atm}\) \(P_{\mathrm{SO}_3} = 2(0.154) = 0.308 \ \mathrm{atm}\) The equilibrium partial pressures are: - \(P_{\mathrm{SO}_{2}} \approx 0.192 \ \mathrm{atm}\) - \(P_{\mathrm{O}_{2}} \approx 0.346 \ \mathrm{atm}\) - \(P_{\mathrm{SO}_{3}} \approx 0.308 \ \mathrm{atm}\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equilibrium Constant
An equilibrium constant (\(K_\mathrm{p}\) for gas-phase reactions or \(K_\mathrm{c}\) for concentration) serves as a measure of the ratio of the concentrations or partial pressures of products and reactants at equilibrium. It provides valuable insight into the extent of a reaction and whether the products or reactants are favored. For the reaction: \[2 \mathrm{SO}_{2}(g) + \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)\]The equilibrium constant expression relates to the given partial pressures as:\[K_{\mathrm{p}} = \frac{P_{\mathrm{SO}_3}^2}{(P_{\mathrm{O}_2})(P_{\mathrm{SO}_2})^2}\]In this specific example, \(K_\mathrm{p} = 0.25\) \(1100 \mathrm{K} \), indicating the relative proportion of product and reactant partial pressures.If \(K_\mathrm{p} > 1\), products are favored at equilibrium; if \(K_\mathrm{p} < 1\), reactants dominate. The constant doesn’t depend on concentrations but on the nature and conditions (temperature, etc.) of the reaction. This defines the equilibrium position without requiring any specifics about the kinetics involved.
Partial Pressure
Partial pressure is simply the pressure that a gas in a mixture would exert if it occupied the entire volume by itself.In chemical equilibrium, knowing the partial pressures of reactants and products is key to predicting reaction direction and extent. This is evident in the equilibrium expression of gas-phase reactions like this one.Initially, partial pressures were \(P_{\mathrm{SO}_2} = P_{\mathrm{O}_2} = 0.50 \mathrm{atm}\) and \(P_{\mathrm{SO}_3} = 0 \mathrm{atm}\).Changes in partial pressure due to the reaction are described as:
  • Decrease in \(\mathrm{SO}_2\): \( -2x \)
  • Decrease in \(\mathrm{O}_2\): \( -x \)
  • Increase in \(\mathrm{SO}_3\): \( +2x \)
Equations are derived to determine equilibrium partial pressures:
  • \(P_{\mathrm{SO}_2} = 0.50 - 2x\)
  • \(P_{\mathrm{O}_2} = 0.50 - x\)
  • \(P_{\mathrm{SO}_3} = 2x\)
With calculations, partial pressures at equilibrium can tell us how much of each species remains or forms.
Reaction Kinetics
Reaction kinetics focus on the rates of chemical reactions and how different conditions affect these rates. Unlike equilibrium, which tells us about the concentrations of reactants and products at equilibrium, kinetics describes how quickly a reaction reaches equilibrium.In our example, although we are given initial partial pressures and \(K_\mathrm{p}\), it is not about how fast equilibrium is reached but about the equilibrium itself. Kinetics would offer details on what path and timeframe are involved.The principle of kinetics would be important if we needed to understand reaction mechanisms or rate laws. Yet, in this context of determining equilibrium partial pressures, we focus more on the concept of equilibrium rather than how the reaction progresses over time.Understanding reaction kinetics though, can be crucial in real-world applications when reactions need to reach equilibrium faster via catalysts or other means. However, to calculate equilibrium partial pressures, these kinetic details remain secondary.

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Most popular questions from this chapter

A sample of iron(II) sulfate was heated in an evacuated container to 920 K, where the following reactions occurred: $$\begin{array}{c}{2 \mathrm{FeSO}_{4}(s) \Longrightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+\mathrm{SO}_{3}(g)+\mathrm{SO}_{2}(g)} \\\ {\mathrm{SO}_{3}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g)}\end{array}$$ After equilibrium was reached, the total pressure was 0.836 atm and the partial pressure of oxygen was 0.0275 atm. Calculate \(K_{\mathrm{p}}\) for each of these reactions.

The synthesis of ammonia gas from nitrogen gas and hydrogen gas represents a classic case in which a knowledge of kinetics and equilibrium was used to make a desired chemical reaction economically feasible. Explain how each of the following conditions helps to maximize the yield of ammonia. a. running the reaction at an elevated temperature b. removing the ammonia from the reaction mixture as it forms c. using a catalyst d. running the reaction at high pressure

At a particular temperature a \(2.00-\mathrm{L}\) flask at equilibrium contains \(2.80 \times 10^{-4}\) mole of \(\mathrm{N}_{2}, 2.50 \times 10^{-5}\) mole of \(\mathrm{O}_{2},\) and \(2.00 \times 10^{-2}\) mole of \(\mathrm{N}_{2} \mathrm{O}\) . Calculate \(K\) at this temperature for the reaction $$2 \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{N}_{2} \mathrm{O}(g)$$ If \(\left[\mathrm{N}_{2}\right]=2.00 \times 10^{-4} M,\left[\mathrm{N}_{2} \mathrm{O}\right]=0.200 M,\) and \(\left[\mathrm{O}_{2}\right]=\) \(0.00245 M,\) does this represent a system at equilibrium?

At a particular temperature, \(K_{\mathrm{p}}=0.25\) for the reaction $$\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)$$ a. A flask containing only \(\mathrm{N}_{2} \mathrm{O}_{4}\) at an initial pressure of 4.5 \(\mathrm{atm}\) is allowed to reach equilibrium. Calculate the equilibrium partial pressures of the gases. b. A flask containing only \(\mathrm{NO}_{2}\) at an initial pressure of 9.0 \(\mathrm{atm}\) is allowed to reach equilibrium. Calculate the equilibrium partial pressures of the gases. c. From your answers to parts a and b, does it matter from which direction an equilibrium position is reached?

Write expressions for \(K\) and \(K_{\mathrm{p}}\) for the following reactions. a. \(2 \mathrm{NH}_{3}(g)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{CH}_{4} \mathrm{O}(s)+\mathrm{H}_{2} \mathrm{O}(g)\) b. \(2 \mathrm{NBr}_{3}(s) \Longrightarrow \mathrm{N}_{2}(g)+3 \mathrm{Br}_{2}(g)\) c. \(2 \mathrm{KClO}_{3}(s) \Longrightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g)\) d. \(\mathrm{CuO}(s)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{Cu}(l)+\mathrm{H}_{2} \mathrm{O}(g)\)

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