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Chemistry: A Molecular Approach | 5th Edition | ISBN: 9780134874371 | Authors: Nivaldo J. Tro ISBN: 9780134874371 2047

Solution for problem 66 Chapter 19

Chemistry: A Molecular Approach | 5th Edition

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Chemistry: A Molecular Approach | 5th Edition | ISBN: 9780134874371 | Authors: Nivaldo J. Tro

Chemistry: A Molecular Approach | 5th Edition

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Problem 66

Consider the reaction:

\(\mathrm{CaCO}_{3}(s) \longrightarrow \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)\)

Estimate \(\Delta G^{\circ}\) for this reaction at each temperature and predict whether or not the reaction is spontaneous. (Assume that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not change too much within the given temperature range.)

a. 298 K

b. 1055 K

c. 1455 K

Text Transcription:

CaCO_3(s) longrightarrow CaO(s) + CO_2( g)

Delta G^circ

Delta H^circ

Delta S^circ

Step-by-Step Solution:
Step 1 of 3

Phase Transitions: 1. Sublimation- solid directly to gas 2. Vaporization- liquid to gas 3. Fusion- solid to liquid 4. Deposition- gas directly to solid 5. Condensation- gas to liquid 6. Freezing- liquid to solid Delta H of sublimation = Delta H of fusion + Delta H of vaporization Equation for Heat required during a phase transition- q = n x Delta H, where n = # of moles Equation for heat required during a single phase- q = c x m x change in temperature, where c = specific heat and m = mass Phase Diagrams- indicates the pressure and temperatures at which the solid, liquid, and gaseous phases of a material are stable. Critical Point- the temperature at above which the gas cannot be made to liquefy, no matter how great the applied pressure Triple Point- the temperature and pressure at which a material exists in the solid, liquid, and gaseous phases simultaneously Example: How much heat is required to vaporize 50g of water at its normal boiling point and subsequently heat the vapor to 250 Degrees Celsius 1. Vaporize 50g H O2 Delta H of vaporization = 40.7 KJ/mol 1molH 2O 50gH 2Ox 18.2g = 2.78 mol H20 q = (2.78mol)(40.7 KJ/mol) = 113 KJ 2. Heat to 250 o m = 50g Delta o = 150 c = 2o03 J/g C q = (2.03 J/g C)(50g)(150 C) = 15200 J 3. Total Energy 15200 J + 113000 J = 128,200 J = 128 KJ

Step 2 of 3

Chapter 19, Problem 66 is Solved
Step 3 of 3

Textbook: Chemistry: A Molecular Approach
Edition: 5
Author: Nivaldo J. Tro
ISBN: 9780134874371

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