INTRODUCTORY CHEMISTRY
8th Edition
ISBN: 2818000163285
Author: CORWIN
Publisher: PEARSON CO
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Chapter 16, Problem 18E
Interpretation Introduction
(a)
Interpretation:
The equilibrium constant expression for the given
Concept introduction:
Interpretation Introduction
(b)
Interpretation:
The equilibrium constant expression for the given chemical reaction is to be stated.
Concept introduction:
Chemical equilibrium is a state in which forward rate of reaction is equal to reverse rate of reaction at a given temperature. Law of chemical equilibrium states that at a given temperature ratio of molar concentration of products, each raised to a power equal to its coefficient, to the molar concentration of reactants, each raised to their coefficient, is constant.
Interpretation Introduction
(c)
Interpretation:
The equilibrium constant expression for the given chemical reaction is to be stated.
Concept introduction:
Chemical equilibrium is a state in which forward rate of reaction is equal to reverse rate of reaction at a given temperature. Law of chemical equilibrium states that at a given temperature ratio of molar concentration of products, each raised to a power equal to its coefficient, to the molar concentration of reactants, each raised to their coefficient, is constant.
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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.
R lactam or lactone considering as weak acid or weak base and why
81. a. Propose a mechanism for the following reaction:
OH
CH2=CHCHC=N
b. What is the product of the following reaction?
HO
H₂O
N=CCH2CH2CH
OH
HO
CH3CCH=CH2
H₂O
C=N
82. Unlike a phosphonium ylide that reacts with an aldehyde or a ketone to form an alkene a sulfonium ulia
Chapter 16 Solutions
INTRODUCTORY CHEMISTRY
Ch. 16 - Prob. 1CECh. 16 - Prob. 2CECh. 16 - Prob. 3CECh. 16 - Prob. 4CECh. 16 - Prob. 5CECh. 16 - Prob. 6CECh. 16 - Prob. 7CECh. 16 - Prob. 8CECh. 16 - Prob. 9CECh. 16 - Prob. 10CE
Ch. 16 - Prob. 1KTCh. 16 - Prob. 2KTCh. 16 - Prob. 3KTCh. 16 - Prob. 4KTCh. 16 - Prob. 5KTCh. 16 - Prob. 6KTCh. 16 - Prob. 7KTCh. 16 - Prob. 8KTCh. 16 - Prob. 9KTCh. 16 - Prob. 10KTCh. 16 - Prob. 11KTCh. 16 - Prob. 12KTCh. 16 - Prob. 13KTCh. 16 - Prob. 14KTCh. 16 - Prob. 15KTCh. 16 - Prob. 16KTCh. 16 - Prob. 17KTCh. 16 - Prob. 18KTCh. 16 - Prob. 1ECh. 16 - Prob. 2ECh. 16 - Prob. 3ECh. 16 - Prob. 4ECh. 16 - Prob. 5ECh. 16 - Prob. 6ECh. 16 - Prob. 7ECh. 16 - Prob. 8ECh. 16 - Prob. 9ECh. 16 - Prob. 10ECh. 16 - Prob. 11ECh. 16 - Prob. 12ECh. 16 - Prob. 13ECh. 16 - Prob. 14ECh. 16 - Prob. 15ECh. 16 - Prob. 16ECh. 16 - Prob. 17ECh. 16 - Prob. 18ECh. 16 - Prob. 19ECh. 16 - Prob. 20ECh. 16 - Prob. 21ECh. 16 - Prob. 22ECh. 16 - Prob. 23ECh. 16 - Prob. 24ECh. 16 - Prob. 25ECh. 16 - Prob. 26ECh. 16 - Prob. 27ECh. 16 - Prob. 28ECh. 16 - Prob. 29ECh. 16 - Prob. 30ECh. 16 - Prob. 31ECh. 16 - Prob. 32ECh. 16 - Prob. 33ECh. 16 - Prob. 34ECh. 16 - Prob. 35ECh. 16 - Prob. 36ECh. 16 - Prob. 37ECh. 16 - Prob. 38ECh. 16 - Prob. 39ECh. 16 - Prob. 40ECh. 16 - Prob. 41ECh. 16 - Prob. 42ECh. 16 - Prob. 43ECh. 16 - Prob. 44ECh. 16 - Prob. 45ECh. 16 - Prob. 46ECh. 16 - Prob. 47ECh. 16 - Prob. 48ECh. 16 - Prob. 1STCh. 16 - Prob. 2STCh. 16 - Prob. 3STCh. 16 - Prob. 4STCh. 16 - Prob. 5STCh. 16 - Prob. 6STCh. 16 - Prob. 7STCh. 16 - Prob. 8STCh. 16 - Prob. 9STCh. 16 - Prob. 10STCh. 16 - Prob. 11STCh. 16 - Prob. 12STCh. 16 - Prob. 13STCh. 16 - Prob. 14STCh. 16 - Prob. 15ST
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