Chemistry [hardcover]
Chemistry [hardcover]
5th Edition
ISBN: 9780393264845
Author: Geoffery Davies
Publisher: NORTON
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Chapter 17, Problem 17.105AP

(a)

Interpretation Introduction

Interpretation: The signs of ΔS and ΔH for the given process are to be predicted. The validation of the fact that the dicarboxylic acid will reform when the melted material cools is to be stated with an explanation.

Concept introduction: Entropy is a thermodynamic property that provides a measure of the randomness of the atoms in the system. Free energy is equal to the difference between the enthalpy and the product of entropy at absolute temperature.

To determine: The signs of ΔS and ΔH for the given process.

(b)

Interpretation Introduction

To determine: If the dicarboxylic acid will reform when the melted material cools.

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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 + N2O5 → NO2 + NO2 + NO2 (K3) Give the expression for the acceptable rate. (A). d[N₂O] dt = -1 2k,k₂[N205] k₁+k₂ d[N₂O5] (B). dt =-k₁[N₂O₂] + k₁[NO2][NO3] - k₂[NO2]³ (C). d[N₂O] dt =-k₁[N₂O] + k₁[N205] - K3 [NO] [N205] (D). d[N2O5] =-k₁[NO] - K3[NO] [N₂05] dt
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.

Chapter 17 Solutions

Chemistry [hardcover]

Ch. 17 - Prob. 17.3VPCh. 17 - Prob. 17.4VPCh. 17 - Prob. 17.5VPCh. 17 - Prob. 17.6VPCh. 17 - Prob. 17.7VPCh. 17 - Prob. 17.8VPCh. 17 - Prob. 17.9QPCh. 17 - Prob. 17.10QPCh. 17 - Prob. 17.11QPCh. 17 - Prob. 17.12QPCh. 17 - Prob. 17.13QPCh. 17 - Prob. 17.14QPCh. 17 - Prob. 17.15QPCh. 17 - Prob. 17.16QPCh. 17 - Prob. 17.17QPCh. 17 - Prob. 17.18QPCh. 17 - Prob. 17.19QPCh. 17 - Prob. 17.20QPCh. 17 - Prob. 17.21QPCh. 17 - Prob. 17.22QPCh. 17 - Prob. 17.23QPCh. 17 - Prob. 17.24QPCh. 17 - Prob. 17.25QPCh. 17 - Prob. 17.26QPCh. 17 - Prob. 17.27QPCh. 17 - Prob. 17.28QPCh. 17 - Prob. 17.29QPCh. 17 - Prob. 17.30QPCh. 17 - Prob. 17.31QPCh. 17 - Prob. 17.32QPCh. 17 - Prob. 17.33QPCh. 17 - Prob. 17.34QPCh. 17 - Prob. 17.35QPCh. 17 - Prob. 17.36QPCh. 17 - Prob. 17.37QPCh. 17 - Prob. 17.38QPCh. 17 - Prob. 17.39QPCh. 17 - Prob. 17.40QPCh. 17 - Prob. 17.41QPCh. 17 - Prob. 17.42QPCh. 17 - Prob. 17.43QPCh. 17 - Prob. 17.44QPCh. 17 - Prob. 17.45QPCh. 17 - Prob. 17.46QPCh. 17 - Prob. 17.47QPCh. 17 - Prob. 17.48QPCh. 17 - Prob. 17.49QPCh. 17 - Prob. 17.50QPCh. 17 - Prob. 17.51QPCh. 17 - Prob. 17.52QPCh. 17 - Prob. 17.53QPCh. 17 - Prob. 17.54QPCh. 17 - Prob. 17.55QPCh. 17 - Prob. 17.56QPCh. 17 - Prob. 17.57QPCh. 17 - Prob. 17.58QPCh. 17 - Prob. 17.59QPCh. 17 - Prob. 17.60QPCh. 17 - Prob. 17.61QPCh. 17 - Prob. 17.62QPCh. 17 - Prob. 17.63QPCh. 17 - Prob. 17.64QPCh. 17 - Prob. 17.65QPCh. 17 - Prob. 17.66QPCh. 17 - Prob. 17.67QPCh. 17 - Prob. 17.68QPCh. 17 - Prob. 17.69QPCh. 17 - Prob. 17.70QPCh. 17 - Prob. 17.71QPCh. 17 - Prob. 17.72QPCh. 17 - Prob. 17.73QPCh. 17 - Prob. 17.74QPCh. 17 - Prob. 17.75QPCh. 17 - Prob. 17.76QPCh. 17 - Prob. 17.77QPCh. 17 - Prob. 17.78QPCh. 17 - Prob. 17.79QPCh. 17 - Prob. 17.80QPCh. 17 - Prob. 17.81QPCh. 17 - Prob. 17.82QPCh. 17 - Prob. 17.83QPCh. 17 - Prob. 17.84QPCh. 17 - Prob. 17.85QPCh. 17 - Prob. 17.86QPCh. 17 - Prob. 17.87QPCh. 17 - Prob. 17.88APCh. 17 - Prob. 17.89APCh. 17 - Prob. 17.90APCh. 17 - Prob. 17.91APCh. 17 - Prob. 17.92APCh. 17 - Prob. 17.93APCh. 17 - Prob. 17.94APCh. 17 - Prob. 17.95APCh. 17 - Prob. 17.96APCh. 17 - Prob. 17.97APCh. 17 - Prob. 17.98APCh. 17 - Prob. 17.99APCh. 17 - Prob. 17.100APCh. 17 - Prob. 17.101APCh. 17 - Prob. 17.102APCh. 17 - Prob. 17.103APCh. 17 - Prob. 17.104APCh. 17 - Prob. 17.105AP
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