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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?
Chemistry
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
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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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