Chemistry, 03.12.2020 17:00 liluv5062

Ethanol (C2H5OH) melts at –114 °C and boils at 78 °C. The enthalpy of fusion of ethanol is 5.02 kJ/mol, and its enthalpy of vaporization is 38.56 kJ/mol. The specific heats of solid and liquid ethanol are 0.97 J/g-K and 2.3 J/g-K, respectively. The average specific heat of gaseous ethanol is about 1.80 J/g-K. a. How much heat is required to convert 35.0 g of ethanol at 27 °C to the vapor phase at 120 °C? b. How much heat is required to convert the same amount of ethanol at –120 °C to the vapor phase at 120 °C?

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Answers

Answer from: kimho2324

First question

$Q = 36826 \ J$

Second question

$Q = 52299.7 \ J$

Explanation:

From the question we are told that

The melting point of Ethanol is T_m = -114 ^oC

The boiling point of Ethanol is T_b = 78^ oC

The enthalpy of fusion of Ethanol is $F = 5.02 \ kJ / mol = 5.02 *10^{3}\ kJ / mol$

The enthalpy of vaporization of Ethanol is $L = 38.56 \ kJ / mol = 38.56 *10^{3} \ J / mol$

The specific heat of solid Ethanol is $c_e = 0.97 \ J/ g \cdot K$

The specific heat of liquid Ethanol is $c_l = 2.3 \ J / g \cdot K$

The mass of the Ethanol given is $m = 35.0 \ g$

Considering the first question

The initial temperature is T_i = 27^oC

The final temperature is T_f = 120^oC

Generally the heat required too raise the Ethanol to its boiling point is mathematically represented as

Q_1 = m * c_l * (T_b - T_i)

=> Q_1 = 35.0 * 2.3 * ( 78 - 27)

=> $Q_1 =4106 \ J$

Genially the number of moles of Ethanol given is mathematically represented as

$n = \frac{m}{Z}$

Here Z is the molar mass of Ethanol with value Z = 46 g/mol

$n = \frac{35}{46 }$

=> $n = 0.7609 \ mol$

Generally the heat of vaporization of the Ethanol is mathematically represented as

Q_2 = n * L

=> $Q_2 =0.7809 * 38.56 * 10^{3}$

=> $Q_2 =29339 \ J$

Generally the heat required too raise the Ethanol from its boiling point to T_f is mathematically represented as

Q_3 = m * c_l * (T_f - T_b)

=> Q_3 = 35 * 2.3 * (120 - 78 )

=> $Q_3 = 3381 \ J$

Generally the total heat required is

Q = Q_1 + Q_2 + Q_3

=> Q = 4106 + 29339 + 3381

=> $Q = 36826 \ J$

Considering the second question

The initial temperature is T_i = -120^oC

The final temperature is T_f = 120^oC

Generally the heat required too raise the Ethanol to its melting point is mathematically represented as

Q_1 = m * c_e * (T_m - T_i)

=> Q_1 = 35.0 * 0.97 * ( -114 - (- 120) )

=> $Q_1 = 203.7 \ J$

Generally the heat of fusion of the Ethanol is mathematically represented as

Q_2 = n * F

=> $Q_2 =0.7809 * 5.02 *10^{3}$

=> $Q_2 =3920 \ J$

Generally the heat required too raise the Ethanol to its boiling point is mathematically represented as

Q_3 = m * c_l * (T_b - T_m)

=> Q_3 = 35.0 * 2.3 * ( 78 - (- 114) )

=> $Q_3 =15456 \ J$

Generally the heat of vaporization of the Ethanol is mathematically represented as

Q_4 = n * L

=> $Q_4 =0.7809 * 38.56 * 10^{3}$

=> $Q_4 =29339 \ J$

Generally the heat required too raise the Ethanol from its boiling point to T_f is mathematically represented as

Q_5 = m * c_l * (T_f - T_b)

=> Q_5 = 35 * 2.3 * (120 - 78 )

=> $Q_5 = 3381 \ J$

Generally the total heat required is

Q = Q_1 + Q_2 + Q_3+Q_4 + Q_5

=> Q = 203.7 + 3920 + 15456 +29339+3381

=> $Q = 52299.7 \ J$

Answer from: Quest

The correct answer is b

Answer from: Quest

each column of the table is called a group. elements in a group share similar properties. groups are read from top to bottom. metal. metalloid. nonmetal. solid. liquid. gas. fe. hg. o. 2. 3. 4. 5. 6. 7. 8. 9. 2. beryllium. 9.012. 4. sodium. 22.990. 11. 3. magnesium. 24.305. 12. potassium. 39.098. 19. 4. calcium. 40.078. 20. i guess i

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Ethanol (C2H5OH) melts at –114 °C and boils at 78 °C. The enthalpy of fusion of ethanol is 5.02 kJ/m...