Given the potential diagram for a reaction:
1 answer:
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Answer : The value of change in entropy for vaporization of water is
Explanation :
Formula used :
where,
= change in entropy of vaporization = ?
= change in enthalpy of vaporization = 40.7 kJ/mol
= boiling point temperature of water =
Now put all the given values in the above formula, we get:
Therefore, the value of change in entropy for vaporization of water is
Answer:
D) 4
Explanation:
If we want to find the asymmetric centers, we must first remember that in an asymmetric center, in the carbon, we will have 4 different types of groups bonded to it.
Therefore we must look for carbons that have 4 different groups. With this in mind, we can look at each carbon.
<u>Carbon 1</u>
In this carbon, we have two bonds with the oxygen, therefore, we dont have 4 different groups. So, this is not an asymmetric center.
<u>Carbon 2</u>
In this carbon, we have an "OH" in the top, an hydrogen in the bottom a carbonyl group in the right, and a CHOH in the left. We have 4 different groups. So, this is an asymmetric center.
<u>Carbon 3</u>
In this carbon, we have an "OH" at the bottom, an hydrogen in the top, a CHOH group in the right (that is bonded to a carbonyl group), and a CHOH in the left. We have 4 different groups. So, this is an asymmetric center.
<u>Carbon 4</u>
In this carbon, we have an "OH" at the top, an hydrogen in the bottom a CHOH group in the right (that is bonded to a CHOH group), and a CHOH in the left. We have 4 different groups. So, this is an asymmetric center.
<u>Carbon 5</u>
In this carbon, we have an "OH" at the bottom, an hydrogen in the top, a CHOH group in the right (that is bonded to a CHOH), and a CH2OH in the left. We have 4 different groups. So, this is an asymmetric center.
<u>Carbon 6</u>
In this carbon, we have two bonds with hydrogens, therefore, we dont have 4 different groups. So, this is not an asymmetric center.
See figure 1
I hope it helps!
