For the part circled in white, I don't understand when simplifying this why we don't divide the coefficient in front of octane by 2. I thought it would be 1/2 Since we divided the other ones by 2.
Electronic Effects
The effect of electrons that are located in the chemical bonds within the atoms of the molecule is termed an electronic effect. The electronic effect is also explained as the effect through which the reactivity of the compound in one portion is controlled by the electron repulsion or attraction producing in another portion of the molecule.
Drawing Resonance Forms
In organic chemistry, resonance may be a mental exercise that illustrates the delocalization of electrons inside molecules within the valence bond theory of octet bonding. It entails creating several Lewis structures that, when combined, reflect the molecule's entire electronic structure. One Lewis diagram cannot explain the bonding (lone pair, double bond, octet) elaborately. A hybrid describes a combination of possible resonance structures that represents the entire delocalization of electrons within the molecule.
Using Molecular Structure To Predict Equilibrium
Equilibrium does not always imply an equal presence of reactants and products. This signifies that the reaction reaches a point when reactant and product quantities remain constant as the rate of forward and backward reaction is the same. Molecular structures of various compounds can help in predicting equilibrium.
For the part circled in white, I don't understand when simplifying this why we don't divide the coefficient in front of octane by 2. I thought it would be 1/2 Since we divided the other ones by 2.
![In the second part, calculate AHn from the
AH's of the reactants and products.
(2)
AH xn
Txn
ΔΗ's
ΔΗ Σn,ΔΗ (products) - Σn,ΔΗ?(reactants)
AHTXN
In the third part, convert from kilojoules of
energy to moles of octane using the conver-
sion factor found in step 2, and then convert
from moles of octane to mass of octane using
(3)
kJ
mol CgH18
g C8H18
kg C3H1
114.22 g C8H18
1 kg
the molar mass.
Conversion factor to be
mol C3H18
1000 g
determined from steps 1 and 2
RELATIONSHIPS USED
molar mass C3H18 = 114.22 g/mol
1 kg = 1000 g
SOLVE Begin by writing the balanced equa-
JLUTION STEP 1
tion for the combustion of octane. For con-
C3H18(1) +
25
O2(8) 8 CO2(8) + 9 H2O(g)
venience, do not clear the 25/2 fraction in
order to keep the coefficient on octane as 1.
SOLUTION STEP 2
Reactant or product
AH; (kJ/mol, from Appendix IIB)
CgH18(/)
-250.1
O2(g)
0.0
Look up (in Appendix IIB) the standard
enthalpy of formation for each reactant and
product and then calculate AHxn:
CO2(g)
-393.5
H20(g)
-241.8
-Continued on the nex](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F61b75115-d70f-4fe2-af93-2076876ad69a%2Fcb2e73f5-b62b-4fd2-b38d-952924db003e%2Fja97pyq_processed.jpeg&w=3840&q=75)
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