The value of K p and K c at 25 0 C for the equilibrium H 2 O(l) ⇌ H 2 O(g) should be calculated. Concept introduction: Equilibrium constant expression for gas phase reactions are written using partial pressures as follows: aA(g) ⇌ bB(g) + cC(g) Or, K P = ( P B ) b ( P C ) c ( P A ) a Here, P A , P B and P C are partial pressures of A, B and C gases respectively. The concentration or partial pressure of pure solids and pure liquids are not included in the equilibrium constant expression. The relationship between K p and K c is given by the following equation: K p = K c ( R T ) Δ n Here, K p = equilibrium constant in terms of partial pressure K c = equilibrium constant in terms of concentrations. R = Universal gas constant T = absolute temperature Δ n = moles of gaseous products - moles of gaseous reactants
The value of K p and K c at 25 0 C for the equilibrium H 2 O(l) ⇌ H 2 O(g) should be calculated. Concept introduction: Equilibrium constant expression for gas phase reactions are written using partial pressures as follows: aA(g) ⇌ bB(g) + cC(g) Or, K P = ( P B ) b ( P C ) c ( P A ) a Here, P A , P B and P C are partial pressures of A, B and C gases respectively. The concentration or partial pressure of pure solids and pure liquids are not included in the equilibrium constant expression. The relationship between K p and K c is given by the following equation: K p = K c ( R T ) Δ n Here, K p = equilibrium constant in terms of partial pressure K c = equilibrium constant in terms of concentrations. R = Universal gas constant T = absolute temperature Δ n = moles of gaseous products - moles of gaseous reactants
Solution Summary: The author explains that the equilibrium constant expression for gas phase reactions is written using partial pressures. The concentration of pure solids and pure liquids is not included in the equation.
16. The proton NMR spectral information shown in this problem is for a compound with formula
CioH,N. Expansions are shown for the region from 8.7 to 7.0 ppm. The normal carbon-13 spec-
tral results, including DEPT-135 and DEPT-90 results, are tabulated:
7
J
Normal Carbon
DEPT-135
DEPT-90
19 ppm
Positive
No peak
122
Positive
Positive
cus
и
124
Positive
Positive
126
Positive
Positive
128
No peak
No peak
4°
129
Positive
Positive
130
Positive
Positive
(144
No peak
No peak
148
No peak
No peak
150
Positive
Positive
してし
3. Propose a synthesis for the following transformation. Do not draw an arrow-pushing
mechanism below, but make sure to draw the product of each proposed step (3 points).
+ En
CN
CN
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Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell