3. The number of vacancies N,in a crystal can be calculated using the Arrhenius equation:Ny= Ne EvIkTwhere Nis total number of atoms, Evis the activation energy for vacancy formation, and kis the Boltzmann's constant and Tis the temperature.Boltzmann constant k = 8.62 x 10S ev/K%3Da) What are the units of T?b) What are the units for E?c) Under what conditions can a crystal have no vacancies?d) Calculate the fraction of atom sites that are vacant for lead and thence the percentageof vacancies for lead at its melting temperature of 327°C ( 600K). Assume an energyfor vacancy formation of 0.55 eV/atom

Usually units are placed in SI system but when you have to evaluate the answer in a particular unit…

3. The number of vacancies Nin a crystal can be calculated using the Arrhenius equation: Ev /kT Ny= Ne where Nis total number of atoms, Ev is the activation energy for vacancy formation, and k is the Boltzmann's constant and Tis the temperature. Boltzmann constant k = 8.62 x 105 ev/K a) What are the units of T b) What are the units for E2 c) Under what conditions can a crystal have no vacancies?

3. The number of vacancies Nin a crystal can be calculated using the Arrhenius equation: Ev /kT Ny= Ne where Nis total number of atoms, Ev is the activation energy for vacancy formation, and k is the Boltzmann's constant and Tis the temperature. Boltzmann constant k = 8.62 x 105 ev/K a) What are the units of T b) What are the units for E2 c) Under what conditions can a crystal have no vacancies?

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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4) Synthesis of ammonia by the Haber process occurs by the reaction N22) + 3H2g) 5 2NH3(g) Species AH (kJ mol"') 4S° (J mol- K-') H2(g) 0. 130.7 N2(8) 0. 191.6 NH3(8) -46.1 192.5 Assuming that AH° and AS° are essentially unchanged in the temperature ranging from 25 °C to 400 °C: a. Calculate K at 25 °C. b. Calculate K at 400 °C. (R= 8.314 J mol- K')
AH°; in kJ/mol N2H«() 39 N2Oalg) 20.39 N20(g) 107.9 H20(g) -257.3 Using the information given above, find AHn for the following reaction: ons N2H4 () + N204 (g) -------> 2 N2O (g) + 2 H20 (g) es
10) Which of the following processes have a AS > 0? А) СНЗОН() — B) N2(g) + 3 H2(g) → 2 NH3(g) C) CH4(g) + H2O(g) → CO(g) + 3 H2(g) D) Na2CO3(s) + H2O(g) + CO2(g) E) All of the above processes have a AS > 0. CH3OH(s) 2 NaHCO3(s)
At 4000C and 350 bar, a 1:3 mixture of nitrogen and hydrogen gases react to form an equilibrium a mixture containing ammonia at a mole fraction of 0.75. Assuming perfect gas behavior, calculate the equilibrium constant K for: N2(g) + 3 H2(g) ⇄ 2 NH3(g)
3. Calculate the number of atoms per cubic meter for metal X at 830 C. The energy for vacancy formation is 2.9 eV/atom; the atomic weight and density for Metal X are 87.3 g/mol and 10 g/cm3, respectively. Answer in Scientific Notation example 1.3e+28. 4. Calculate the equilibrium number of vacancies per cubic meter for metal X at 544C. The energy for vacancy formation is 0.5 eV/atom; the atomic weight and density for Metal X are 46.5 g/mol and 5 g/cm3, respectively. Answer in scientific notation. Round your answer to 2 significant figures.
Vanadium tetrachloride (VCl4 ) is a bright red colored liquid with a vapor pressure of 5x10-2 torr at 298K. Briefly discuss why the standard state of VCl4 is a liquid and VCl2 is an ionic solid? How many microstates are in the gas phase V4+ free ion ground term and the V4+ free ion ground J state ?
02 Question Calculate the value of K, for the reaction 2N₂(g) + O₂(g) 2N₂O(g) 1 at 298.15 K and 1773 K. Thermodynamic data for N₂O(g) are: AH", - 82.05 kJ/mol; S-219.9 J/mol K: AG-104.2 kJ/mol. 1st attempt Part 1 298.15 K See page 695 Jd See Periodic Table See Hint ANAWER
A[CH,0] For the reaction below, what is with respect to At At 2 C2Н4(9) + 2 Оз(9) — 4СH20(9) + 02(9) O AICH,0] At At O 4[CH,0] 2A[0 3] At At A[CH,0] At 4At O 4ICH,0] Δε At O 4[CH,0] At 2At
Thermo - Problem 14.44 NO is a well-known example of a molecular system in which excited electronic energy levels are readily accessible at room temperature. Both the ground and excited electronic states are Part A doubly degenerate and are separated by 121.1 cm-1. (Note: the Boltzmann constant k = 1.38 x 10-34 J.s, speed of the light c = 3.00 · 1010 cm · s-1 the Planck constant h = 6.626 · 1034 J.s). Evaluate the electronic partition function for this molecule at 373 K. Express the electronic partition function to three significant figures. ? qE = Submit Request Answer Part B Determine the temperature at which qp = 3. Express the temperature in kelvins to three significant figures. ? T = K Submit Request Answer
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for the following the A 6° volue Calculote reac 4) 298 K. reaction running +1/2 O2 1g) Es CO2 g) value constant Assumeng that the AH° reac tempero ture studied, colculo te the in the range A6° reac value at S00 K, CS CamScanner ile tarandı
9:24 Consider the reaction at 298 K. 2Cgraphite) + 0,g)- 2COg) AH=-2210 klmol Calculate the quantities for AS. AS. ASmir AS J(mol - K) AS M(mol - K) J(mol - K)