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marin [14]
3 years ago
9

Explain how hydrogen bonding contributes to water's high heat of vaporization

Chemistry
1 answer:
pashok25 [27]3 years ago
4 0

The heat/enthalpy of vaporization of water represents the energy input required to convert one mole of water into vapor at a constant temperature. Intermolecular forces including hydrogen bondings of significant strength hold water molecules in place under its liquid state. Whereas the molecules experience almost no intermolecular interactions under the gaseous state- consider the way noble gases molecules interact. It is thus necessary to supply sufficient energy to overcome all intermolecular interactions present in the substance under its liquid state to convert the substance into a gas. The heat of vaporization is thus related to the strength of the intermolecular interactions.

Water molecules contain hydrogen atoms bonded directly to oxygen atoms. Oxygen atoms are highly electronegative and take major control of electrons in hydrogen-oxygen bonds. Hydrogen atoms in water molecules thus experience a strong partial-positive charge and would attract lone pairs of electron on neighboring water molecules. "Hydrogen bonds" refer to the attraction between hydrogen atoms bonded to electronegative elements and lone pairs of electrons. The hydrogen-oxygen bonds in water molecules are so polarized that hydrogen bonds in water are stronger than both dipole-dipole interactions and London Dispersion Forces in most other molecules. It thus take high amounts of energy to separate water molecules sufficiently apart such that they no longer experience intermolecular interactions and behave collectively like a gas. As a result, water has one of the highest heat of vaporization among covalent molecules of similar sizes.

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The reaction between ethylene and hydrogen bromide to form ethyl bromide is carried out in a continuous reactor. The product str
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Explanation:

The first thing to do is to write out the chemical equation showing the reaction between the two chemical compounds.

C2H4 + HBr ------------------------> C2H5Br.

The chemical species/compounds reacted in the ratio of 1 : 1 : 1. That is one mole of C2H4 reacted with one mole of HBr to give one mole of C2H5Br.

So, we are given; C2H5Br = 58.7 mol%, HBr = 18.3 mol%. Thus, for C2H4, we have [ 100 - (58.7 + 18.3)] = 23 mol%. Therefore, the limiting reagent is HBr.

So, if we have 100 mol/s ;

C2H5Br = 100 × 0.587 = 58.7 mol/s, HBr = 100 × 0.183 = 18.3 mol/s and C2H4 = 100 × 0.23 = 23 mol/s.

Therefore, the total value of the flow rate in the feed = (HBr in feed) mol/s + (C2H4 in feed) mol/s = (58.7 + 18.3)mol/s + (58.7 + 23) mol/s =( 77 + 81.7) mol/s = 158.7 mol/s. .

So, 100/158.7 × 30.5 = 192.2.

Also the feed composition in HBr and C2H4 is;

HBr = 77/158.7 = 0.4852, C2H4 = 81.7/158.7 = 0.515.

Therefore, the number of HBr in feed and C2H4 in feed = 160m/s and 200m/s respectively (approx.)

Therefore, the fraction conversion is;

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During the reaction of 2,4-dinitrofluorobenzene with methoxide in methanol it was observed that it occurs at a significantly hig
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Explanation:

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If a pork roast must absorb kJ to fully cook, and if only 10% of the heat produced by the barbecue is actually absorbed by the r
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Answer:

1,033.56 grams of carbon dioxide was emitted into the atmosphere.

Explanation:

C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(g),\Delta H_{rxn}=-2044 kJ/mol (assuming)

Energy absorbed by pork,E = 1.6\times 10^3 kJ (assuming)

Total energy produced by barbecue = Q

Percentage of energy absorbed by pork = 10%

10\%=\frac{E}{Q}\times 100

Q=\frac{E}{10}\times 100=\frac{1.6\times 10^3 kJ}{10}\times 100=1.6\times 10^4 kJ

Since, it is a energy produced in order to indicate the direction of heat produced we will use negative sign.

Q = -1.6\times 10^4 kJ

Moles of propane burnt to produce Q energy =n

n\times \Delta H_{rxn}=Q

n=\frac{Q}{\Delta H_{rxn}}=\frac{-1.6\times 10^4 kJ}{-2044 kJ/mol}=7.83 mol

According to reaction , 1 mol of propane gives 3 moles of carbon dioxide. then 7.83 moles of will give:

\frac{1}{3}\times 7.83 mol=23.49 mol carbon dioxide gas.

Mass of 23.49 moles of carbon dioxide gas:

23.49 mol × 44 g/mol =1,033.56 g

1,033.56 grams of carbon dioxide was emitted into the atmosphere.

8 0
3 years ago
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