At \(2200^{\circ} \mathrm{C}, K_{\mathrm{p}}=0.050\) for the reaction $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) $$ What is the partial pressure of \(\mathrm{NO}\) in equilibrium with \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) that were placed in a flask at initial pressures of \(0.80\) and \(0.20\) atm, respectively?

Short Answer

Expert verified
The partial pressure of NO at equilibrium can be found using the ICE table and the provided Kp value. After setting up the balanced reaction equation, Kp expression, and ICE table, the equation \(0.050 = \frac{(2x)^2}{(0.800-x)(0.200-x)}\) can be derived. Solving this quadratic equation for x yields two possible solutions; however, only the positive value of x is physically possible. Using the positive value of x, the equilibrium partial pressure of NO is found by calculating \(2x\), giving the final answer.

Step by step solution

01

1. Write down the balanced equation and Kp expression

The balanced equation for the reaction is: \[N_{2}(g) + O_{2}(g) \rightleftharpoons 2NO(g)\] The Kp expression for this reaction is: \[K_p = \frac{[NO]^2}{[N_{2}][O_{2}]}\] Where [NO], [N2], and [O2] represent the partial pressures of NO, N2, and O2 at equilibrium, and Kp = 0.050.
02

2. Create an ICE table to list initial, change, and equilibrium pressures

An ICE (Initial, Change, Equilibrium) table is a useful way to keep track of the changes and equilibrium values in a chemical reaction. | | N2 | O2 | NO | |------|--------|-----|----| | I | 0.800 | 0.20 | 0 | | C | -x | -x | +2x | | E | 0.800-x | 0.20-x | 2x | Here, x is the change in partial pressures when the equilibrium is established.
03

3. Substitute equilibrium pressures into the Kp expression

Substitute the equilibrium values from the ICE table into the Kp expression: \[K_p = \frac{(2x)^2}{(0.800-x)(0.200-x)}\] Plug in the given value of Kp: \[0.050 = \frac{(2x)^2}{(0.800-x)(0.200-x)}\]
04

4. Solve the equation for x

To solve for x, you can either use quadratic formula or try to simplify the equation. First, let's multiply both sides by (0.800-x)(0.200-x) to get rid of the fraction: \[0.050(0.800-x)(0.200-x)= (2x)^2\] Now expand both side of the equation: \[0.010 - 0.210x + x^2 = 4x^2\] Rearrange the equation to form a quadratic equation: \[3x^2 + 0.210x - 0.010 = 0\] Use the quadratic formula to solve for x: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] Here, a = 3, b = 0.210, and c = -0.010. Plugging in the values, we get two possible solutions for x. However, one of them yields a negative value for partial pressures, which is physically impossible. So, we take the positive value of x.
05

5. Calculate the partial pressure of NO at equilibrium

We found the value of x, which represents the changes in partial pressures. To find the equilibrium partial pressure of NO, we need to use the value we found for x in the ICE table's equilibrium row: For NO, the equilibrium partial pressure is 2x. Plug in the positive value of x and calculate the partial pressure of NO. This will be the final answer to the problem.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Equilibrium Constant
Understanding the equilibrium constant is crucial for grasping how chemical reactions reach a state of balance. The equilibrium constant, represented by the symbol K, is a number that provides a measure of where the equilibrium lies for a given chemical reaction. For gas-phase reactions, the equilibrium constant is often given as Kp, which is based on the partial pressures of gases involved.

Using the equation from the exercise, \[K_p = \frac{[NO]^2}{[N_{2}][O_{2}]}\], Kp relates to how product and reactant pressures influence the reaction's balance. A small Kp value (0.050 in this case) suggests that, at equilibrium, the reactants (N2 and O2) will predominate over the products (NO) at the specified temperature of 2200°C. In other words, there will be a higher concentration of unreacted N2 and O2 than that of NO. Understanding this concept allows students to predict the direction of the shift in a reaction when changes occur in the system.
ICE Table Method
The ICE table is a systematic approach that helps understand changes that occur as reactants move toward equilibrium to form products. ICE stands for Initial, Change, and Equilibrium, representing different stages of the reaction.

In the case of our problem, the initial pressures of N2 and O2 are given, while NO is initially not present. As the reaction proceeds towards equilibrium, the pressures of N2 and O2 decrease while that of NO increases. The changes in pressures are represented by a variable 'x' in the ICE table. This methodical layout aids in visualizing the reaction progression and simplifies the process of solving for the equilibrium constant, Kp. Students are encouraged to use this method to avoid confusion and ensure that all changes in the reactant and product concentrations are accounted for correctly when determining the equilibrium state.
Partial Pressure
Partial pressure, an essential aspect of chemical equilibrium in gaseous systems, refers to the pressure exerted by each individual gas in a mixture. According to Dalton's law, the total pressure exerted by the mixture is the sum of the partial pressures of each gas composing it. When dealing with chemical equilibria involving gases, partial pressures play a pivotal role.

In our exercise, the partial pressure of NO at equilibrium is of particular interest and is determined by the change in moles of gas 'x' during the reaction from the initial pressures of N2 and O2. Upon establishing the value of 'x', it can be substituted back into the ICE table to calculate the equilibrium partial pressures of each gas. Understanding the concept of partial pressure is instrumental in solving problems related to gas-phase equilibrium, as it intimately links to the equilibrium constant expression.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Consider the reaction \(\mathrm{A}(g)+2 \mathrm{~B}(g) \rightleftharpoons \mathrm{C}(g)+\mathrm{D}(g)\) in a 1.0-L rigid flask. Answer the following questions for each situation \((\mathrm{a}-\mathrm{d})\) : i. Estimate a range (as small as possible) for the requested substance. For example, [A] could be between \(95 M\) and \(100 M\) ii. Explain how you decided on the limits for the estimated range. iii. Indicate what other information would enable you to narrow your estimated range. iv. Compare the estimated concentrations for a through d, and explain any differences. a. If at equilibrium \([\mathrm{A}]=1 M\), and then \(1 \mathrm{~mol} \mathrm{C}\) is added, estimate the value for \([\mathrm{A}]\) once equilibrium is reestablished. b. If at equilibrium \([\mathrm{B}]=1 M\), and then \(1 \mathrm{~mol} \mathrm{C}\) is added, estimate the value for [B] once equilibrium is reestablished. c. If at equilibrium \([\mathrm{C}]=1 M\), and then \(1 \mathrm{~mol} \mathrm{C}\) is added, estimate the value for \([\mathrm{C}]\) once equilibrium is reestablished. d. If at equilibrium \([\mathrm{D}]=1 M\), and then \(1 \mathrm{~mol} \mathrm{C}\) is added, estimate the value for [D] once equilibrium is reestablished.

Suppose the reaction system $$ \mathrm{UO}_{2}(s)+4 \mathrm{HF}(g) \rightleftharpoons \mathrm{UF}_{4}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) $$ has already reached equilibrium. Predict the effect that each of the following changes will have on the equilibrium position. Tell whether the equilibrium will shift to the right, will shift to the left, or will not be affected. a. Additional \(\mathrm{UO}_{2}(s)\) is added to the system. b. The reaction is performed in a glass reaction vessel; \(\mathrm{HF}(g)\) attacks and reacts with glass. c. Water vapor is removed.

The compound \(\mathrm{VCl}_{4}\) undergoes dimerization in solution: $$ 2 \mathrm{VCl}_{4} \rightleftharpoons \mathrm{V}_{2} \mathrm{Cl}_{8} $$ When \(6.6834 \mathrm{~g} \mathrm{VCl}_{4}\) is dissolved in \(100.0 \mathrm{~g}\) carbon tetrachloride, the freezing point is lowered by \(5.97^{\circ} \mathrm{C}\). Calculate the value of the equilibrium constant for the dimerization of \(\mathrm{VCl}_{4}\) at this temperature. (The density of the equilibrium mixture is \(1.696\) \(\mathrm{g} / \mathrm{cm}^{3}\), and \(K_{\mathrm{f}}=29.8^{\circ} \mathrm{C} \mathrm{kg} / \mathrm{mol}\) for \(\mathrm{CCl}_{4} .\).

At a particular temperature, \(K=4.0 \times 10^{-7}\) for the reaction $$ \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) $$ In an experiment, \(1.0 \mathrm{~mol} \mathrm{~N}_{2} \mathrm{O}_{4}\) is placed in a 10.0-L vessel. Calculate the concentrations of \(\mathrm{N}_{2} \mathrm{O}_{4}\) and \(\mathrm{NO}_{2}\) when this reaction reaches equilibrium.

Methanol, a common laboratory solvent, poses a threat of blindness or death if consumed in sufficient amounts. Once in the body, the substance is oxidized to produce formaldehyde (embalming fluid) and eventually formic acid. Both of these substances are also toxic in varying levels. The equilibrium between methanol and formaldehyde can be described as follows: $$ \mathrm{CH}_{3} \mathrm{OH}(a q) \rightleftharpoons \mathrm{H}_{2} \mathrm{CO}(a q)+\mathrm{H}_{2}(a q) $$ Assuming the value of \(K\) for this reaction is \(3.7 \times 10^{-10}\), what are the equilibrium concentrations of each species if you start with a \(1.24 M\) solution of methanol? What will happen to the concentration of methanol as the formaldehyde is further converted to formic acid?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free