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(a)
Interpretation:
The change in standard Gibbs free energy of reaction needs to be calculated for the given reaction at 298 K.
Concept Introduction:
The change in standard Gibbs free energy of a reaction can be calculated using the change in Gibbs free energy of formation of reactant and products from the
The formula used is as follows:
(b)
Interpretation: The change in standard Gibbs free energy of reaction needs to be calculated for the given reaction at 298 K.
Concept Introduction:
The change in standard Gibbs free energy of a reaction can be calculated using the change in Gibbs free energy of formation of reactant and products from the thermodynamic table (at 298 K).
The formula used is as follows:
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Chapter 20 Solutions
CHEMISTRY:MOLECULAR...(LL)-W/CONNECT
- Redraw the flowchartarrow_forwardredraw the flowchart with boxes and molecules written in themarrow_forwardPart I. a) Elucidate the structure of compound A using the following information. • mass spectrum: m+ = 102, m/2=57 312=29 • IR spectrum: 1002.5 % TRANSMITTANCE Ngg 50 40 30 20 90 80 70 60 MICRONS 5 8 9 10 12 13 14 15 16 19 1740 cm M 10 0 4000 3600 3200 2800 2400 2000 1800 1600 13 • CNMR 'H -NMR Peak 8 ppm (H) Integration multiplicity a 1.5 (3H) triplet b 1.3 1.5 (3H) triplet C 2.3 1 (2H) quartet d 4.1 1 (2H) quartet & ppm (c) 10 15 28 60 177 (C=0) b) Elucidate the structure of compound B using the following information 13C/DEPT NMR 150.9 MHz IIL 1400 WAVENUMBERS (CM-1) DEPT-90 DEPT-135 85 80 75 70 65 60 55 50 45 40 35 30 25 20 ppm 1200 1000 800 600 400arrow_forward
- • Part II. a) Elucidate The structure of compound c w/ molecular formula C10 11202 and the following data below: • IR spectra % TRANSMITTANCE 1002.5 90 80 70 60 50 40 30 20 10 0 4000 3600 3200 2800 2400 2000 1800 1600 • Information from 'HAMR MICRONS 8 9 10 11 14 15 16 19 25 1400 WAVENUMBERS (CM-1) 1200 1000 800 600 400 peak 8 ppm Integration multiplicity a 2.1 1.5 (3H) Singlet b 3.6 1 (2H) singlet с 3.8 1.5 (3H) Singlet d 6.8 1(2H) doublet 7.1 1(2H) doublet Information from 13C-nmR Normal carbon 29ppm Dept 135 Dept -90 + NO peak NO peak 50 ppm 55 ppm + NO peak 114 ppm t 126 ppm No peak NO peak 130 ppm t + 159 ppm No peak NO peak 207 ppm по реак NO peakarrow_forwardCould you redraw these and also explain how to solve them for me pleasarrow_forwardNonearrow_forward
- Draw the curved-arrow mechanism with the drawings of the molecules, not just abbreviations. -NO₂ Sn, HCl (aq) E D H (CH3CO)₂O -NH2 CH3arrow_forwardWhat is/are the product(s) of the following reaction? Select all that apply. * HI A B C OD OH A B OH D Carrow_forwardIn the image, the light blue sphere represents a mole of hydrogen atoms, the purple or teal spheres represent a mole of a conjugate base. A light blue sphere by itself is H+. Assuming there is 2.00 L of solution, answer the following: The Ka of the left & right solution is? The pH of the left & right solution is? The acid on the left & right is what kind of acid?arrow_forward
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