Equilibrium: Le Chatelier's Principle and Q vs. K
13 Questions
0 Views

Choose a study mode

Play Quiz
Study Flashcards
Spaced Repetition
Chat to Lesson

Podcast

Play an AI-generated podcast conversation about this lesson

Questions and Answers

Explain how the reaction quotient, Q, helps predict the direction a reversible reaction will shift to reach equilibrium, and relate this to the equilibrium constant, K.

Q predicts the shift. If Q < K, the reaction shifts right (towards products). If Q > K, it shifts left (towards reactants). If Q = K, the reaction is at equilibrium.

For the reaction $N_2(g) + 3H_2(g) ightleftharpoons 2NH_3(g)$, how would increasing the total pressure of the system at equilibrium affect the partial pressure of $NH_3$? Explain using Le Chatelier's principle.

Increasing the total pressure will shift the equilibrium towards the side with fewer moles of gas, which is the product side ($NH_3$). Therefore, the partial pressure of $NH_3$ will increase.

Briefly describe how a catalyst affects the equilibrium of a reversible reaction. Does it change the equilibrium constant or the rates of the forward and reverse reactions?

A catalyst does not change the equilibrium constant. It increases the rates of both the forward and reverse reactions equally, allowing equilibrium to be reached faster.

Consider the endothermic reaction: $A(g) ightleftharpoons B(g)$. If the temperature of the system at equilibrium is decreased, what will happen to the concentration of A? Explain your answer.

<p>Decreasing the temperature will shift the equilibrium to favor the reverse reaction because the reverse reaction generates heat. Therefore, the concentration of A will increase.</p> Signup and view all the answers

For the equilibrium reaction $2SO_2(g) + O_2(g) ightleftharpoons 2SO_3(g)$, the enthalpy change, $\Delta H$, is negative. How will increasing the temperature affect the value of the equilibrium constant, K?

<p>Since the reaction is exothermic ($\Delta H$ is negative), increasing the temperature will shift the equilibrium to favor the reverse reaction, decreasing the value of K.</p> Signup and view all the answers

Explain how the relative intensities of peaks in a mass spectrum can be used to determine the isotopic abundance of an element.

<p>The intensity of each peak is directly proportional to the abundance of the corresponding isotope. By comparing the intensities, we can calculate the relative abundance.</p> Signup and view all the answers

Describe the key differences between bonding and antibonding molecular orbitals in terms of electron density and energy levels.

<p>Bonding orbitals have increased electron density between nuclei and lower energy. Antibonding orbitals have a node between nuclei and higher energy.</p> Signup and view all the answers

How does increasing the number of nodes in a molecular orbital affect its energy, and why?

<p>More nodes result in higher energy because nodes indicate fewer electrons between the nuclei, increasing repulsion and kinetic energy of the electrons.</p> Signup and view all the answers

Explain why homonuclear diatomic molecules of the second period (Li to Ne) do not all exist as stable diatomic species. Use molecular orbital theory.

<p>The stability depends on the bond order. If the number of electrons in bonding orbitals is insufficient to outweigh those in antibonding orbitals (bond order of zero), the molecule is unstable.</p> Signup and view all the answers

Describe the relationship between bond order, bond length, and bond energy. How are they related?

<p>As bond order increases, bond length decreases, and bond energy increases. Higher bond order means stronger, shorter bonds requiring more energy to break.</p> Signup and view all the answers

Explain the concept of resonance and why it is used to represent certain molecules. Provide an example of a molecule that requires resonance structures.

<p>Resonance is used when a single Lewis structure cannot accurately represent electron distribution. It involves multiple structures with electrons delocalized. Ozone ($O_3$) is an example.</p> Signup and view all the answers

How does electronegativity difference between two atoms influence the polarity of a bond? Give an example.

<p>Greater electronegativity difference leads to a more polar bond, with a partial positive charge ($\delta+$) on the less electronegative atom and a partial negative charge ($\delta-$) on the more electronegative atom. For example, in H-Cl, Cl is more electronegative, so the bond is polar.</p> Signup and view all the answers

What are the differences between sigma ($\sigma$) and pi ($\pi$) bonds in terms of their formation and electron density distribution?

<p>Sigma bonds are formed by end-to-end overlap with electron density along the internuclear axis. Pi bonds are formed by sideways overlap with electron density above and below the internuclear axis.</p> Signup and view all the answers

Flashcards

Exam Point Total

A chemistry exam worth a total of 100 points.

Exam Identification

The first exam of Chemistry 112A.

Student ID

To be written on each page of the exam.

Exam Duration

  1. 5 hours
Signup and view all the flashcards

Number of questions

8 questions

Signup and view all the flashcards

The exam contains

10 numbered pages, including the last SCRATCH PAGE.

Signup and view all the flashcards

Student ID

To write your student ID on each page of the exam

Signup and view all the flashcards

6-digit WUSTL ID

9 9 9 9 9 9

Signup and view all the flashcards

Chemical Equilibrium

A state where the rate of forward and reverse reactions are equal, and the net change in concentrations of reactants and products is zero.

Signup and view all the flashcards

Equilibrium Constant (K)

The ratio of product pressures to reactant pressures at equilibrium, each raised to the power of their stoichiometric coefficients.

Signup and view all the flashcards

Method of Successive Approximations (MOSA)

A method to simplify equilibrium calculations by assuming that the change in concentration ('x') is negligible compared to initial concentrations.

Signup and view all the flashcards

Partial Pressure

The pressure exerted by an individual gas in a mixture of gases.

Signup and view all the flashcards

ICE Table

A table used to calculate equilibrium concentrations by tracking Initial concentrations, Change in concentrations, and Equilibrium concentrations.

Signup and view all the flashcards

Study Notes

  • Exam consists of questions covering representative topics
  • Exam is on chemistry
  • Professor Bleeke is teaching
  • Exam is worth 100 points

Exam Key Details

  • The alotted time for the exam is 1.5 hours
  • Exam includes 8 questions on 10 numbered pages
  • Students are asked to write their student ID on each page of the exam
  • The exam also includes a scratch page

Question 1a

  • Considers the balanced chemical reaction of 2 ClO (g) = Cl (g) + ClO2 (g)
  • K = 4.26 × 10−4 at 385 K
  • 1.240 atm of ClO (g) and 0.330 atm of Cl (g) are sealed in a container at 385 K
  • No ClO2 (g) is present initially
  • Partial pressures of all three gases, ClO (g), Cl (g) and ClO2 (g) (in atm), after equilibrium is established
  • Pclo = 1.236 atm
  • Pc1= 0.332 atm
  • Pc102= 0.00196 atm

Question 1b

  • Considers the balanced chemical reaction: BrO (g) + ClO (g) = ClO2 (g) + Br (g)
  • The equilibrium constant, K, for this reaction at 385 K is 0.040
  • The reaction is endothermic
  • Plot A depicts the reaction spontaneously proceeding in the forward direction to reach equilibrium at a temperature greater than 385 K
  • Plot B depicts the reaction spontaneously proceeding in the forward direction to reach equilibrium at a temperature greater than 385 K
  • Plot C depicts the reaction spontaneously proceeding in the reverse direction to reach equilibrium at a temperature equal to 385 K

Question 2

  • Considers generic, balanced chemical reactions that are either endothermic or exothermic in the forward direction
  • Each reaction is allowed to reach equilibrium
  • Reactions are heated from T1, to T2
  • Reaction shifts spontaneously in either the forward or the reverse direction to re-establish equilibrium.
  • A) A(g) + 2 B(g) = 2 C(g) is endothermic proceeds in the forward direction
  • B) A(g) + B(g) = 2 C(g) is exothermic proceeds in the reverse direction
  • C) A(g) + 2 B(g) = C(g) + D(g) exothermic is ambiguous
  • D) A(g) + B(g) = C(s) + 2 D(g) endothermic proceeds in the forward direction

Question 3

  • The gas phase reaction of 4 NO (g) + 6 H2O (g) = 4 NH3 (g) + 5 O2 (g)
  • K = 8.32 × 10-19 at temperature T
  • 1.25 atm of NH3(g) and 1.25 atm of O2(g) are placed in a sealed container at temperature T
  • No NO (g) or H2O (g) are initially present
  • PNH3= 0.251 atm
  • Po₂= 0.00119 atm

Question 4a

  • The solid compound (NH4)2CO3 decomposes according to the reaction (NH4)2CO3 (s) = 2 NH3 (g) + CO2 (g) + H2O (g)
  • Total pressure is 3.52 atm after equilibrium is established after the excess (NH4)2CO3 (s) is sealed in an evacuated container at 750 K
  • K = 2.40

Question 4b

  • The chemical reaction is at temperature T = 775 K
  • The equilibrium constant K at 775 K is equal to 5.47
  • When 1.40 atm of each gas are initially sealed in a container at 775 K, a calculation shows that the the Q (3.84) is less than K (5.47), which makes the reaction is spontaneous in the forward direction and no solid forms.
  • Minimum of 0.27 atm of NH3 must be added

Question 5a

  • A 0.40 M aqueous solution of HF is prepared at 25°C
  • Ka (HF, 25°C) = 6.6 × 10-4
  • pKa (HF, 25°C) = 3.18
  • The percent dissociation is 3.98%

Question 5b

  • Determine the number of moles of HClO4 that should be added to 600.0 mL of pure water at 25°C
  • Solution should have same final pH as the 0.40 M aqueous solution of HF
  • 0.00955 moles should be added

Question 5c

  • Comparing a 600.0 mL sample of the 0.40 M aqueous solution of HF and the 600.0 mL sample of the HClO4 solution
  • The [H3O⁺] is the same for both solutions after equilibrium is established for each solution
  • The same number of moles of OH- would be required to completely neutralize each solution.
  • Both solutions would have a pH of 7.00 after completely neutralizing the acids with OH

Question 6a

  • Ka for the weak acid, hypochlorous acid, HClO, is 3.0 ×_10-8 (pKa=7.53) at 25°C
  • The mass of solid potassium hypochlorite, K+ClO¯, that must be added to a 500 mL aqueous solution of 0.10 M HClO to prepare a buffer solution with pH = 8.00 is 13.4 grams

Question 6b

  • The equilibrium chemical equation is ClO (aq) + H2O (l) = HClO (aq) + OH¯ (aq)

Question 6c

  • 50 mL of a 0.20 M aqueous HCl solution is added to the buffer (from part a)
  • The pH = 7.89

Question 6d

  • The number of moles of KClO that must be added to the buffer solution in part (c) to bring it back to the original pH of 8.00 is 0.0391 mol

Question 7a

  • 200.0 mL aqueous solution at 25°C contains a weak, monoprotic acid, HA
  • Titrated with a 0.50 M aqueous solution of NaOH at 25°C
  • 0.076 moles HA

Question 7b

  • 0.076 moles of HA is converted to A- after adding of 152 mL NaOH
  • Kb = 5.32 × 10-10

Question 7c

  • The monoprotic weak acid, HA from the titration is HN3
  • Ka = 1.9 × 10-5

Question 8a

  • 500.0 mL of a 0.50 M aqueous solution of MgCl2 is added to 250.0 mL of a 0.50 M aqueous solution of NaF at 25° C
  • A precipitate of the sparingly-soluble salt, MgF2, is formed
  • MgCl2 and NaF are completely soluble in water to produce Mg2+, Cl-, Na+, and F− ions
  • Ksp for MgF2 = 6.6 × 10-9 at 25°C
  • MW (MgF2) = 62.30 g/mol
  • 3.89 grams of MgF2 precipitates

Question 8b

  • The mass of MgF2 that re-dissolves is 3.79 × 10-3 grams MgF2

Studying That Suits You

Use AI to generate personalized quizzes and flashcards to suit your learning preferences.

Quiz Team

Related Documents

Chemistry 112A Exam 1 Key PDF

Description

Explore chemical equilibrium, focusing on Le Chatelier's principle, the reaction quotient Q, and the equilibrium constant K. Learn how pressure, temperature, and catalysts affect equilibrium position and constant values determining reaction direction.

More Like This

Use Quizgecko on...
Browser
Browser