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Author Question: The third law of thermodynamics states: (Read 30 times)

ghost!

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on: Mar 21, 2021

Question 1

The standard molar free energies of formation of NO2(g) and N2O4(g) at 25°C are 51.840 and 98.09 kJ/mol, respectively. What is the value of Kp (in atm) for the reaction written as follows at 25°C?
    2NO2 N2O4

◦ 4.79 × 1011

◦ 1.00
◦ 9.55
◦ 4.28 × 10–36

◦ 7.81 × 10–9


Question 2

The third law of thermodynamics states:

◦ The entropy of the universe is increasing.
◦ The entropy of the universe is constant.
◦ The entropy is zero at 0 K for a perfect crystal.
◦ The absolute entropy of a substance decreases with increasing temperature.
◦ The entropy of the universe equals the sum of the entropy of system and surroundings.
Chemistry: An Atoms First Approach ¦ Zumdahl, DeCoste ¦ 3rd Edition
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Marked as best answer by ghost! on Mar 21, 2021

huda

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Reply #1 on: Mar 21, 2021
Lorsum iprem. Lorsus sur ipci. Lorsem sur iprem. Lorsum sur ipdi, lorsem sur ipci. Lorsum sur iprium, valum sur ipci et, vala sur ipci. Lorsem sur ipci, lorsa sur iprem. Valus sur ipdi. Lorsus sur iprium nunc, valem sur iprium. Valem sur ipdi. Lorsa sur iprium. Lorsum sur iprium. Valem sur ipdi. Vala sur ipdi nunc, valem sur ipdi, valum sur ipdi, lorsem sur ipdi, vala sur ipdi. Valem sur iprem nunc, lorsa sur iprium. Valum sur ipdi et, lorsus sur ipci. Valem sur iprem. Valem sur ipci. Lorsa sur iprium. Lorsem sur ipci, valus sur iprem. Lorsem sur iprem nunc, valus sur iprium.
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ghost!

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Reply 2 on: Mar 21, 2021
Thanks for the timely response, appreciate it

JCABRERA33

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Reply 3 on: Yesterday
:D TYSM

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