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What amount of heat (in kJ) is required to convert 14.0 g of an unknown liquid (MM = 67.44 g/mol) at 43.5 °C to a gas at 128.2 °C? (specific heat capacity of liquid = 1.18 J/g･°C; specific heat capacity of gas = 0.792 J/g･°C; ∆Hvap = 30.1 kJ/mol; normal boiling point, Tb = 97.4°C)

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Answer:

1.24 kJ is required to convert 14 g of liquid from 43.5°C to 128.2°C

Explanation:

This is a typical calorimetry problem:

We have to assume, no heat is lost to sourrounding.

First of all, we need to go from 43.5°C to 97.4°C, the boiling point.

Q = Ce . m . ΔT

We replace data, 1.18° J/g . 14 g . (97.4°C - 43.5°C)

Heat for the first stage is: 890.4 Joules

Now we have to change the state, and we need the ΔH. As we do not have latent heat, we can proceed like this:

1 mol release 30.1 kJ at vaporization.

We convert the mass to moles → 14 g.  1mol/ 67.44g = 0.207 mol

0.207 mol will release (0.207 . 30.1 kJ) = 6.25 kJ

Now, we are at gaseous phase.

Q = Ce . m . ΔT → 0.792 J/g°C . 14g . (128.2°C - 97.4°C)

Q = 341.5 Joules

To determine the amount of heat, we sum all the obtained values:

890.4 Joules + 6250 Joules + 341.5 Joules = 1238.2 J

We convert to kJ →  1238.2 J . 1kJ / 1000J = 1.24 kJ

Dsdrajlinidn

The heat required to convert 14.0 g of an unknown liquid at 43.5 °C to gas at 128.2 °C is 7.48 kJ.

We want to calculate the heat required to convert 14.0 g of an unknown liquid at 43.5 °C to gas at 128.2 °C.

We can divide this process in 3 steps.

  1. Heating of the liquid from 43.5 °C to 97.4 °C (normal boiling point).
  2. Vaporization of the liquid at 97.4 °C.
  3. Heating of the gas from 97.4 °C to 128.2 °C.

1. Heating of the liquid from 43.5 °C to 97.4 °C

We will calculate the heat for this step (Q₁) using the following expression.

Q₁ = c(l) × m × ΔT

Q₁ = (1.18 J/g･°C) × 14.0 g × (97.4 °C - 43.5 °C) = 890 J = 0.890 kJ

where,

  • c(l) is the specific heat capacity of the liquid.
  • m is the mass of the substance.
  • ΔT is the change in the temperature.

2. Vaporization of the liquid at 97.4 °C.

We will calculate the heat for this step (Q₂) using the following expression.

Q₂ = (m/M) × ΔHvap

Q₂ = [14.0 g/(67.44 g/mol)] × 30.1 kJ/mol = 6.25 kJ

where,

  • m is the mass of the substance.
  • M is the molar mass of the substance.
  • ΔHvap is the enthalpy of vaporization of the substance.

3. Heating of the gas from 97.4 °C to 128.2 °C.

We will calculate the heat for this step (Q₃) using the following expression.

Q₃ = c(g) × m × ΔT

Q₃ = (0.792 J/g･°C) × 14.0 g × (128.2 °C - 97.4 °C) = 342 J = 0.342 kJ

where,

  • c(g) is the specific heat capacity of the gas.
  • m is the mass of the substance.
  • ΔT is the change in the temperature.

4. Total amount of heat required (Q)

Q = Q₁ + Q₂ + Q₃ = 0.890 kJ + 6.25 kJ + 0.342 kJ = 7.48 kJ

The heat required to convert 14.0 g of an unknown liquid at 43.5 °C to gas at 128.2 °C is 7.48 kJ.

Learn more about heating curves here: https://brainly.com/question/10481356

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