Chemistry Q: intermolecular forces, spontaneity of rxns, basicity

[161927]

Here's a few questions from my study materials and I disagree with the provided answers. I'll post the given answers after people have a chance to respond.

Q1: For the molecule shown in the above figure, what will the predominant intermolecular forces be? SHOWN: Formaldehyde (C=O-H2)

A. dipole-dipole forces
B. coordinate covalent bonding forces
C. hydrogen bond forces
D. London forces

Q2: A reaction results in lower enthalpy and higher entropy. Which of the following four statements can be said of this reaction?

A. The reaction will be spontaneous at sufficiently low temperatures.
B. The reaction will be non-spontaneous.
C. The reaction will be spontaneous at sufficiently high temperatures.
D. The reaction will be spontaneous.

Q3: Of the following functional groups, which is alkaline?

A. amines
B. esters
C. amides
D. phenols


Thanks

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[bluemonkey]

Here's a few questions from my study materials and I disagree with the provided answers. I'll post the given answers after people have a chance to respond.

Q1: For the molecule shown in the above figure, what will the predominant intermolecular forces be? SHOWN: Formaldehyde (C=O-H2)

A. dipole-dipole forces
B. coordinate covalent bonding forces
C. hydrogen bond forces
D. London forces

Q2: A reaction results in lower enthalpy and higher entropy. Which of the following four statements can be said of this reaction?

A. The reaction will be spontaneous at sufficiently low temperatures.
B. The reaction will be non-spontaneous.
C. The reaction will be spontaneous at sufficiently high temperatures.
D. The reaction will be spontaneous.

Q3: Of the following functional groups, which is alkaline?

A. amines
B. esters
C. amides
D. phenols


Thanks

just a guess...

 

[161927]

just a guess...

Your answers are in agreement with the book's answer key.

Q1: I now understand why "hydrogen bond forces" is incorrect. There are no acidic hydrogens in this molecule to hydrogen bond to the partially polarized carbonyl oxygen.


Q2: This question is ambiguous worded IMO. What does "results in lower enthalpy" mean? It could have gone from a positive enthalpy to a slightly less positive enthalpy (200-->150 J for example) and still be non-spontaneous.


Q3: How do we avoid choosing phenols? The pKa of a phenolic hydrogen is approximately 9, which would be alkaline...right?

 

[BerkReviewTeach]

Your answers are in agreement with the book's answer key.

Q1: I now understand why "hydrogen bond forces" is incorrect. There are no acidic hydrogens in this molecule to hydrogen bond to the partially polarized carbonyl oxygen.


Q2: This question is ambiguous worded IMO. What does "results in lower enthalpy" mean? It could have gone from a positive enthalpy to a slightly less positive enthalpy (200-->150 J for example) and still be non-spontaneous.


Q3: How do we avoid choosing phenols? The pKa of a phenolic hydrogen is approximately 9, which would be alkaline...right?

Q2: Even from 200 to 150 means that deltaH is a negative number. Greater entropy means that deltaS is a positive number. Plugging into the equation:
deltaG = deltaH - TdeltaS will result in a negative number no matter what T is. A negative deltaG is associated with a spontaneous reaction.

Q3: Alkaline means basic, which implies that when added to water, the pH would increase. Phenol has a pKa of 10.01, so it is weakly acidic. When added to water, it dissociates and generates H+, which results in a slightly acidic solution.

Be careful not to mistake pKa for pH. That is a very common error. Amines are basic, because when they are added to water, the nitrogen will pick up a proton from water and generate hydroxide. Thus the solution becomes basic.

Don't the materials have detailed explanations making these points? The best learning occurs when you review an answer and suddenly get it. That's when it sticks for good. If the practice questions you are using have only one-sentence answer explanations that don't help much, you should work with better practice questions.

 

[JA Prufrock]

regarding question 2, are you familiar with the equation:

dG = dH - TdS

(d stands for delta)

That's how you determine spontaneity in a reaction. If dG is negative, the reaction will proceed spontaneously in the forward direction. The question tells you that you have a lower enthalpy, meaning a negative enthalpy change [dH = H(final) - H(initial)]. It also tells you have a higher entropy (so a positive change in entropy). Let's just use simple numbers and make dH = -100 and dS = +20.

dG = -100 - T(+20)

As you can see, dG is going to be negative, regardless of the temperature (unless temperature is negative, but we're dealing with Kelvins here so that's impossible). Thus, the reaction is spontaneous.

 

[kish180]

"Q1: I now understand why "hydrogen bond forces" is incorrect. There are no acidic hydrogens in this molecule to hydrogen bond to the partially polarized carbonyl oxygen."
It doesnt matter if the hydrogens were acidic or not. For instance, even if there were alpha (acidic) hydrogens, it STILL would not constitute Hbonding because Hbonding needs to have O,N,F ---- H ---- O,N,F

 

[161927]

"Q1: I now understand why "hydrogen bond forces" is incorrect. There are no acidic hydrogens in this molecule to hydrogen bond to the partially polarized carbonyl oxygen."
It doesnt matter if the hydrogens were acidic or not. For instance, even if there were alpha (acidic) hydrogens, it STILL would not constitute Hbonding because Hbonding needs to have O,N,F ---- H ---- O,N,F

The question asks about intermolecular forces not intramolecular forces. If there was an acidic hydrogen, the molecule could form hydrogen bonds: an H from one formaldehyde molecule bonding to the carbonyl oxygen of a nearby formaldehyde.