All categories

3 years ago

PLEASE HELP!!

Chemistry

1 answer:

uranmaximum [27]3 years ago

3 0

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 diagram:

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 heat capacity or thermal capacity is the energy needed to increase the temperature of a certain substance in a unit of temperature. 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 $\frac{1 mol}{200.59 g}$ x 27.983 $\frac{J}{mol K}$ 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. 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. Therefore, to determine the energy necessary for the mercury to pass from gaseous state to liquid, the negative of the enthalpy of vaporization must be taken, 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 $\frac{1 mol}{200.59 g}$ x 59.11 $\frac{kJ}{mol}$

→ Q₂ = 22.10 kJ → Q₂ = 22100 J

Step 3:

The heat in step 3 will be

Q₃ = n Cp ΔT = 75.0 g x $\frac{1 mol}{200.59 g}$ x 27.983 $\frac{J}{mol K}$ 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, the heat lost when 75.0 g of mercury vapor at 650 K condenses to a liquid at 297 K is - 25.8 K

You might be interested in

A. Calculate the empirical formula of a molecule with percent compositions: 55.3% potassium (K), 14.6% phosphorus (P), and 30.1%

Otrada [13]

The way you calculate the empirical formula is to firstly assume 100g. To find each elements moles you take each elements percentage listed, times it by one mole and divide it by its atomic mass. (ex: moles of K =55.3g x 1 mole/39.1g, therefore there is 1.41432225 moles of Potassium) Once you’ve completed this for every element you list each elements symbol beside it’s number of moles and divide by the smallest number because it can only go into its self once. After you’ve done this, you’ve found your empirical formula, which is the simplest whole number ratio of atoms in a compound. I’ve added an example of a empirical question I completed last semester :)

6 0

3 years ago

Read 2 more answers

Does the identity of an atom change if we add or subtract electrons or neutrons? Explain.

AURORKA [14]

Adding or removing neutrons from the nucleus are how isotopes are created. Protons carry a positive electrical charge and they alone determine the charge of the nucleus. Adding or removing protons from the nucleus changes the charge of the nucleus and changes that atom's atomic number.

7 0

3 years ago

How could you determine the acceleration of marble?

arsen [322]

Since f=ma assuming you knew the mass of the marble and the total amount of force acting on it than you would divide the amount of force by the mass.

4 0

3 years ago

How does the formal charge play a role in resonance?

Crank

In a Lewis structure, formal charges can be assigned to each atom by treating each bond as if one-half of the electrons are assigned to each atom. ... Resonance occurs in cases where two or more Lewis structures with identical arrangements of atoms but different distributions of electrons can be written.

3 0

3 years ago

Which step in the free-radical chlorination of methane do you expect to be the most exothermic? attempt this problem without loo

stiks02 [169]

The termination step of the free-radical chlorination of methane is the most stable one among all three steps.

The free-radical substitution reaction between chlorine and methane features three major steps:

Initiation, during which chlorine molecules undergo homolytic fission to produce chlorine free radicals. Ultraviolet radiations are typically applied to supply the energy required for breaking the chlorine-chlorine single bonds. The initiation step is thus endothermic.

Propagation, a process in which chlorine free radicals react with methane molecules and remove a hydrogen atom from the alkane to produce hydrogen chloride and an alkyl radical e.g., $\cdot \text{CH}_3$ . The carbon-containing free radical would react with chlorine molecules to produce chloromethane and yet another chlorine free radical. This process can well repeat itself to chlorinate a significant number of methane molecules.

Termination. Free radicals combine to produce molecules. For example, two chlorine free radicals would combine to produce a chlorine molecule, whereas two alkyl free radicals would combine to produce an alkane with two-carbon atoms in its backbone.

Chemical processes that increase the stability of a substance reduces its chemical potential energy. Energy conserves, thus such processes would also release energy equal to the potential energy lost in quantity. Free radicals are unstable and- as seen in the propagation step- compete readily with neutral molecules for their electrons. The propagation step keeps the number of free radicals constant and is therefore more exothermic than the initiation step. The termination step reduces the number of free radicals, increase the stability of the system by the greatest extent, and is therefore the most exothermic step among the three.

3 0

3 years ago