chloroform (chcl3) has a Hvap of 29.6 kj/mol and its boiling point occurs at 61.7c. the specific heat capacity of liquid chloroform is 0.622 J/gC while that of the gas is 0.714 J/gC. How much energy (in kJ) needs to be removed to condense 250 g of gaseous chloroform from 75.0C to a liquid at 40.0C? Molar mass of chloroform is 119.37 g/mol.
I'm not going to go through this this will all of the temps but you can. Here is how you do the problem.
q1 = heat removed to move chloroform vapor from starting T to boiling point.
q1 = mass CHCl3 x specific heat vapor x (Tfinal-Tinitial)
q2 = heat removed to condense vapor at boiling point to liquid at boiling point.
q2 = mass x heat vap.
q3 = heat removed to move T from boiling point to final T in liquid phase.
q3 = mass CHCl3 x specific heat liquid CHCl3 x (Tfinal-Tinitial)
Total q = q1 + q2 + q3.
Note: Everything is in J/g except for dHvap. I would change that to J/g first thing out of the box.
To find the energy required to condense gaseous chloroform, we need to consider the following steps:
1. Convert the mass of chloroform from grams to moles.
2. Calculate the energy required to cool the gas from 75.0°C to its boiling point (61.7°C).
3. Calculate the energy required to condense the gas at its boiling point.
4. Calculate the energy required to cool the liquid chloroform from its boiling point (61.7°C) to 40.0°C.
Let's begin by calculating the moles of chloroform (CHCl3) using its molar mass.
Step 1: Convert grams to moles.
Number of moles (n) = Mass (m) / Molar mass (M)
Given:
Mass = 250 g
Molar mass = 119.37 g/mol
n = 250 g / 119.37 g/mol
n ≈ 2.0944 mol
Now, let's calculate the energy required to cool the gas from 75.0°C to its boiling point (61.7°C).
Step 2: Calculate the energy using the specific heat capacity (C) formula.
Energy (Q) = Mass (m) × Specific heat capacity (C) × Temperature change (ΔT)
Given:
Mass = 250 g
Specific heat capacity (C_gas) = 0.714 J/g°C
Initial temperature (T_initial) = 75.0°C
Final temperature (T_final) = 61.7°C
ΔT = T_final - T_initial
ΔT = 61.7°C - 75.0°C
ΔT = -13.3°C (Note: The negative sign indicates a decrease in temperature)
Q_gas = 250 g × 0.714 J/g°C × (-13.3°C)
Q_gas ≈ -2374.965 J
Next, let's calculate the energy required to condense the gas at its boiling point.
Step 3: Calculate the energy required to condense.
Energy (Q_condensation) = Number of moles (n) × Enthalpy of vaporization (Hvap)
Given:
Number of moles (n) = 2.0944 mol
Enthalpy of vaporization (Hvap) = 29.6 kJ/mol
Q_condensation = 2.0944 mol × 29.6 kJ/mol
Q_condensation ≈ 61.99744 kJ
Finally, let's calculate the energy required to cool the liquid chloroform from its boiling point (61.7°C) to 40.0°C.
Step 4: Calculate the energy using the specific heat capacity (C) formula.
Energy (Q_liquid) = Mass (m) × Specific heat capacity (C) × Temperature change (ΔT)
Given:
Mass = 250 g
Specific heat capacity (C_liquid) = 0.622 J/g°C
Initial temperature (T_initial) = 61.7°C
Final temperature (T_final) = 40.0°C
ΔT = T_final - T_initial
ΔT = 40.0°C - 61.7°C
ΔT = -21.7°C (Note: The negative sign indicates a decrease in temperature)
Q_liquid = 250 g × 0.622 J/g°C × (-21.7°C)
Q_liquid ≈ -3301.775 J
To find the total energy required, sum up the energies calculated above:
Total energy (Q_total) = Q_gas + Q_condensation + Q_liquid
Q_total ≈ -2374.965 J + 61.99744 kJ - 3301.775 J
Now, we have to convert the unit from joules (J) to kilojoules (kJ). Divide by 1000 to convert J to kJ:
Q_total ≈ (-2374.965 J - 3301.775 J) / 1000 kJ
Q_total ≈ -5676.74 J / 1000 kJ
Q_total ≈ -5.67674 kJ
Therefore, approximately 5.68 kJ of energy needs to be removed to condense 250 g of gaseous chloroform from 75.0°C to a liquid at 40.0°C.