Physical Chemistry Plus Mastering Chemistry With Etext -- Access Card Package (3rd Edition) (engel Physical Chemistry Series)
3rd Edition
ISBN: 9780321766205
Author: Thomas Engel, Philip Reid
Publisher: PEARSON
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Chapter 8, Problem 8.6CP
Interpretation Introduction
Interpretation:The reason of close values of triple point temperature and the normal freezing point for most of the substance needs to be explained.
Concept Introduction: All substances can mainly exist in three phases; solid, liquid and gases. These three phases can convert into each other by the application of temperature and pressure such as by heating of solid, it converts to liquid and then gaseous
Similarly, the gases can be compressed at high pressure to liquefy. The interconversion of three phases of substance can be show on the phase diagram.
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Students have asked these similar questions
A 0.10 M solution of acetic acid (CH3COOH, Ka = 1.8 x 10^-5) is titrated with a 0.0250 M solution of magnesium hydroxide (Mg(OH)2). If 10.0 mL of the acid solution is titrated with 20.0 mL of the base solution, what is the pH of the resulting solution?
For the decomposition reaction of N2O5(g): 2 N2O5(g) → 4 NO2(g) + O2(g), the following mechanism has been proposed:
N2O5 NO2 + NO3 (K1) | NO2 + NO3 → N2O5 (k-1) | NO2 + NO3 NO2 + O2 + NO (k2) | NO + N2O51 NO2 + NO2 + NO2 (K3)
→
Give the expression for the acceptable rate.
→
→
(A).
d[N205]
dt
==
2k,k₂[N₂O₂]
k₁+k₁₂
(B).
d[N2O5]
=-k₁[N₂O] + k₁[NO₂] [NO3] - k₂[NO₂]³
dt
(C).
d[N2O5]
=-k₁[N₂O] + k [NO] - k₂[NO] [NO]
d[N2O5]
(D).
=
dt
= -k₁[N2O5] - k¸[NO][N₂05]
dt
Do not apply the calculations, based on the approximation of the stationary state, to make them perform correctly. Basta discard
the 3 responses that you encounter that are obviously erroneous if you apply the formula to determine the speed of a reaction.
For the decomposition reaction of N2O5(g): 2 N2O5(g) → 4 NO2(g) + O2(g), the following mechanism has been proposed:
N2O5 NO2 + NO3 (K1) | NO2 + NO3 → N2O5 (k-1) | NO2 + NO3 NO2 + O2 + NO (k2) | NO + N2O51 NO2 + NO2 + NO2 (K3)
→
Give the expression for the acceptable rate.
→
→
(A).
d[N205]
dt
==
2k,k₂[N₂O₂]
k₁+k₁₂
(B).
d[N2O5]
=-k₁[N₂O] + k₁[NO₂] [NO3] - k₂[NO₂]³
dt
(C).
d[N2O5]
=-k₁[N₂O] + k [NO] - k₂[NO] [NO]
d[N2O5]
(D).
=
dt
= -k₁[N2O5] - k¸[NO][N₂05]
dt
Do not apply the calculations, based on the approximation of the stationary state, to make them perform correctly. Basta discard
the 3 responses that you encounter that are obviously erroneous if you apply the formula to determine the speed of a reaction.
Chapter 8 Solutions
Physical Chemistry Plus Mastering Chemistry With Etext -- Access Card Package (3rd Edition) (engel Physical Chemistry Series)
Ch. 8 - Prob. 8.1CPCh. 8 - Prob. 8.2CPCh. 8 - Prob. 8.3CPCh. 8 - Prob. 8.4CPCh. 8 - Prob. 8.5CPCh. 8 - Prob. 8.6CPCh. 8 - Prob. 8.7CPCh. 8 - Prob. 8.8CPCh. 8 - Prob. 8.9CPCh. 8 - Prob. 8.10CP
Ch. 8 - Prob. 8.11CPCh. 8 - Prob. 8.12CPCh. 8 - Prob. 8.13CPCh. 8 - Prob. 8.14CPCh. 8 - Prob. 8.15CPCh. 8 - Prob. 8.16CPCh. 8 - Prob. 8.17CPCh. 8 - Prob. 8.18CPCh. 8 - Prob. 8.19CPCh. 8 - Prob. 8.20CPCh. 8 - Prob. 8.21CPCh. 8 - Prob. 8.1NPCh. 8 - Prob. 8.2NPCh. 8 - Prob. 8.3NPCh. 8 - Prob. 8.4NPCh. 8 - Prob. 8.5NPCh. 8 - Prob. 8.6NPCh. 8 - Prob. 8.7NPCh. 8 - Prob. 8.8NPCh. 8 - Prob. 8.9NPCh. 8 - Prob. 8.10NPCh. 8 - Prob. 8.11NPCh. 8 - Prob. 8.12NPCh. 8 - Prob. 8.13NPCh. 8 - Prob. 8.14NPCh. 8 - Prob. 8.15NPCh. 8 - Prob. 8.16NPCh. 8 - Prob. 8.17NPCh. 8 - Prob. 8.18NPCh. 8 - Prob. 8.20NPCh. 8 - Prob. 8.21NPCh. 8 - Prob. 8.22NPCh. 8 - Prob. 8.23NPCh. 8 - Prob. 8.24NPCh. 8 - Prob. 8.25NPCh. 8 - Prob. 8.26NPCh. 8 - Prob. 8.27NPCh. 8 - Prob. 8.28NPCh. 8 - Prob. 8.29NPCh. 8 - Prob. 8.30NPCh. 8 - Prob. 8.31NPCh. 8 - Prob. 8.32NPCh. 8 - Prob. 8.33NPCh. 8 - Prob. 8.34NPCh. 8 - Prob. 8.35NPCh. 8 - Prob. 8.36NPCh. 8 - Prob. 8.37NPCh. 8 - Prob. 8.38NPCh. 8 - Prob. 8.39NPCh. 8 - Prob. 8.40NPCh. 8 - Prob. 8.41NPCh. 8 - Prob. 8.42NPCh. 8 - Prob. 8.43NPCh. 8 - Prob. 8.44NPCh. 8 - Prob. 8.45NP
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