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DanielleElmas [232]

4 years ago

6

Ethanol is not usually a solvent used for sn2 reactions, why not? can you predict an undesirable side reaction that might occur

using the ethanol as a solvent that would not occur if we used thf as a solvent?

1 answer:

Greeley [361]4 years ago

8 0

R-X + C2H5OH ===> R-O-C2H5

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Find an expression for the change in entropy when two blocks of the same substance of equal mass, one at the temperature Th and

Gre4nikov [31]

Explanation:

Relation between entropy change and specific heat is as follows.

$\Delta S = C_{p} log (\frac{T_{2}}{T_{1}})$

The given data is as follows.

mass = 500 g, $C_{p}$ = 24.4 J/mol K

$T_{h}$ = 500 K, $T_{c}$ = 250 K

Mass number of copper = 63.54 g /mol

Number of moles = $\frac{mass}{/text{\molar mass}}$

= $\frac{500}{63.54}$

= 7.86 moles

Now, equating the entropy change for both the substances as follows.

$7.86 \times 24.4 \times [T_{f} - 250]$ = $7.86 \times 24.4 \times [500 -T_{f}]$

$T_{f} - 250 = 500 - T_{f}$

$2T_{f}$ = 750

So, $T_{f}$ = $375^{o}C$

For the metal block A, change in entropy is as follows.

$\Delta S = C_{p} log (\frac{T_{2}}{T_{1}})$

= $24.4 log [\frac{375}{500}]$

= -3.04 J/ K mol

For the block B, change in entropy is as follows.

$\Delta S = C_{p} log (\frac{T_{2}}{T_{1}})$

= $24.4 log [\frac{375}{250}]$

= 4.296 J/Kmol

And, total entropy change will be as follows.

= 4.296 + (-3.04)

= 1.256 J/Kmol

Thus, we can conclude that change in entropy of block A is -3.04 J/ K mol and change in entropy of block B is 4.296 J/Kmol.

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Answer: I think the answer is 27N.

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