Question

Tetrahydrofuran (THF) is a common organic solvent with a boiling point of 339 K. Calculate the total energy (q) required to convert 27.3 g of THF at 298 K to a vapor at 373 K. The specific heat of liquid THF is 1.70 J/g K, the specific heat of THF vapor is 1.06 J/g K, and the heat of vaporization of THF is 444 J/g.

Answer 1

Step-by-step explanation:

For the given reaction, the temperature of liquid will rise from 298 K to 339 K. Hence, heat energy required will be calculated as follows.

`Q_(1) = mC_(1) \Delta T_(1)`

Putting the given values into the above equation as follows.

`Q_(1) = mC_(1) \Delta T_(1)`

=
`27.3 g * 1.70 J/g K * 41`

= 1902.81 J

Now, conversion of liquid to vapor at the boiling point (339 K) is calculated as follows.

`Q_(2)` = energy required =
`mL_(v)`

`L_(v)` = latent heat of vaporization

Therefor, calculate the value of energy required as follows.

`Q_(2)` =
`mL_(v)`

=
`27.3 * 444`

= 12121.2 J

Therefore, rise in temperature of vapor from 339 K to 373 K is calculated as follows.

`Q_(3) = mC_(2) \Delta T_(2)`

Value of
`C_(2)` = 1.06 J/g,
`\Delta T_(2)` = (373 -339) K = 34 K

Hence, putting the given values into the above formula as follows.

`Q_(3) = mC_(2) \Delta T_(2)`

=
`27.3 g * 1.06 J/g * 34 K`

= 983.892 J

Therefore, net heat required will be calculated as follows.

Q =
`Q_(1) + Q_(2) + Q_(3)`

= 1902.81 J + 12121.2 J + 983.892 J

= 15007.902 J

Thus, we can conclude that total energy (q) required to convert 27.3 g of THF at 298 K to a vapor at 373 K is 15007.902 J.