While bond energies and bond enthalpies can be used to estimate the heat of reaction (enthalpy change of a reaction), H, the heat of neutralization is the heat released when 1 mole of water is generated by the reaction of an acid and a base (reaction).
For the same type of bond, bond enthalpies differ from compound to compound. For instance, the C-H bond enthalpy in methane is nearly identical to that of ethane, butane, etc. When we look up the bond enthalpy for a C-H bond in a table of bond enthalpies, the average number that results may only be accurate to two or three significant figures.
Each compound's enthalpies of production are listed, and those numbers take into account any minor variations in the enthalpies of each bond. Therefore, the result will be more accurate if you utilize formation enthalpies rather than average bond enthalpies to compute a given reaction's enthalpy change.
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D the lowest amount of energy
Explanation:
Since, the given reaction is as follows.

Initial: 36.1 atm 0 0
Change: 2x x x
Equilibrium: (36.1 - 2x) x x
Now, expression for
of this reaction is as follows.
![K_{p} = \frac{[N_{2}][O_{2}]}{[NO]^{2}}](https://tex.z-dn.net/?f=K_%7Bp%7D%20%3D%20%5Cfrac%7B%5BN_%7B2%7D%5D%5BO_%7B2%7D%5D%7D%7B%5BNO%5D%5E%7B2%7D%7D)
As the initial pressure of NO is 36.1 atm. Hence, partial pressure of
at equilibrium will be calculated as follows.
![K_{p} = \frac{[N_{2}][O_{2}]}{[NO]^{2}}](https://tex.z-dn.net/?f=K_%7Bp%7D%20%3D%20%5Cfrac%7B%5BN_%7B2%7D%5D%5BO_%7B2%7D%5D%7D%7B%5BNO%5D%5E%7B2%7D%7D)
x = 18.1 atm
Thus, we can conclude that partial pressure of
at equilibrium is 18.1 atm.
The answer is c)11 because copper has 11 valence electrons
Erosion, ocean waves and weathering.