what is the main 'impurity' collected with the oil in distillation. PLEASE HELP IM IN END OF YEAR EXAM. II WILL GIVE BRAINLIEST
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Explanation:
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Answer: - 25.8 kJ
The overall process of interest in the question is the following
Hg (g) (650 K) → Hg (l) (297 K)
However, for mercury to pass from a gaseous state in 650 K to a liquid state in 297 K, it must go through a series of steps:
Step 1. Gaseous mercury at 650 K should cool down to 629.88 K, temperature corresponding to the vaporization temperature of this substance.
Step 2. Gaseous mercury goes to liquid state at 629.88 K .
Step 3. The liquid mercury at 629.88 K is cooled until it reaches 297 K.
This series of steps can be represented through the following <u>diagram</u>:
Hg (g) (650 K) → Hg (g) (629.88 K) → Hg (l) (629.88 K) → Hg (l) (297 K)
(1) (2) (3)
Then the total heat involved in the process will be equal to the sum of the heats inherent to steps 1, 2 and 3. We proceed to calculate the heats for each of the steps.
Step 1:
The heat in step 1 will be given by
Q = n Cp ΔT
where n is the number of moles of mercury, Cp is the heat capacity and ΔT is equal to the difference between the temperatures at the end (T₂) and at the beginning of the process (T₁), that is to say
ΔT = T₂ - T₁
You should know that the <u>heat capacity or thermal capacity is the energy needed to increase the temperature of a certain substance in a unit of temperature.</u> The heat capacity of mercury is Cp = 27.983 J / mol K
Then the heat in step 1 will be,
Q₁ = 75.0 g x x 27.983
x (629.88 K - 650 K)
→ Q₁ = - 210.5 J
Step 2:
In this step a change in the state of the mercury occurs, since it condenses from a gaseous state to a liquid state. In this case the heat involved in the process will be given by ,
Q = - n ΔHvap
where ΔHvap is the enthalpy of vaporization. <u>The enthalpy of vaporization is the amount of energy necessary for the mass unit of a substance that is in equilibrium with its own vapor at a pressure of one atmosphere to pass completely from the liquid state to the gaseous state.</u> Therefore, to determine the energy necessary for the mercury to pass from gaseous state to liquid,<u> the negative of the enthalpy of vaporization must be taken</u>, as it is done in the previous equation with the minus sign that is placed.
The enthalpy of vaporization of mercury is ΔHvap = 59.11 kJ/mol
Then the heat in step 2 will be,
Q₂ = - 75.0 g x x 59.11
→ Q₂ = 22.10 kJ → Q₂ = 22100 J
Step 3:
The heat in step 3 will be
Q₃ = n Cp ΔT = 75.0 g x x 27.983
x (297 K - 629.88 K)
→ Q₃ = - 3483 J
Finally the heat involved in the overall process will be ,
Q = Q₁ + Q₂ + Q₃ = - 210.5 J - 22100 J - 3483 J = - 25794 J
→ Q = - 25.8 kJ
So,<u> the heat lost when 75.0 g of mercury vapor at 650 K condenses to a liquid at 297 K is - 25.8 K</u>