Chemistry, 15.10.2019 22:00 Ashley606hernandez

Select whether there is no work done by the system, work done by the system or work done by the surroundings for each system described below:

a. h2o (l) ⟶ h2o (g) δh = 44.0 kj/mol b. 2 no (g) + o2 (g) ⟶ 2 no2(g) δh = -114.1 kj/mol c. cl (g) + o3 (g) ⟶ clo(g) + o2(g) δh = -163 kj/mol d. caco3 (s) ⟶ cao (s) + co2(g) δh = 110.1 kj/mol e. 4 nh3(g) + 5 o2(g) ⟶ 4 no (g) + 6 h2o(l) δh = -906 kj/mol

Answers: 2

Show answers

Answers

Answer from: keving4three

The systems in which work is done by the system:

$H_2O (l) \rightarrow H_2O (g), \Delta H = 44.0 kJ/mol$

$CaCO_3 (s) \rightarrow CaO (s) + CO_2(g) ,\Delta H = 110.1 kJ/mol$

The systems in which work is done by the surroundings:

$2 NO (g) + O_2 (g) \rightarrow 2 NO_2(g),\Delta H = -114.1 kJ/mol$

$4 NH_3(g) + 5 O_2(g) \rightarrow 4 NO (g) + 6 H_2O(l) ,\Delta H = -906 kJ/mol$

The systems in which no work is done:

$Cl (g) + O_3 (g) \rightarrow ClO(g) + O_2(g) \Delta H = -163 kJ/mol$

Explanation:

When the work is done by the system the energy of the system decreases, and volume increases.When the work is done by the surrounding on the system, the energy of the system increases, and volume decreases.When there is no change in the volume of the system it is concluded as no work is done by the system.

Given:

Following systems:

A. $H_2O (l) \rightarrow H_2O (g), \Delta H = 44.0 kJ/mol$

B. $2 NO (g) + O_2 (g) \rightarrow 2 NO_2(g),\Delta H = -114.1 kJ/mol$

C. $Cl (g) + O_3 (g) \rightarrow ClO(g) + O_2(g) \Delta H = -163 kJ/mol$

D. $CaCO_3 (s) \rightarrow CaO (s) + CO_2(g) ,\Delta H = 110.1 kJ/mol$

E. $4 NH_3(g) + 5 O_2(g) \rightarrow 4 NO (g) + 6 H_2O(l) ,\Delta H = -906 kJ/mol$

To find:

Whether there is no work done by the system, work done by the system or work done by the surroundings for each system

Solution:

A. $H_2O (l) \rightarrow H_2O (g), \Delta H = 44.0 kJ/mol$

The increase in the volume of the system as gaseous moles of product is more than gaseous moles of reactant.

Hence, the work done by the system.

B. $2 NO (g) + O_2 (g) \rightarrow 2 NO_2(g),\Delta H = -114.1 kJ/mol$

The decrease in the volume of the system as gaseous moles of product is less than gaseous moles of reactant.

Hence, the work done by the surroundings.

C. $Cl (g) + O_3 (g) \rightarrow ClO(g) + O_2(g) \Delta H = -163 kJ/mol$

The no change in the volume of the system as gaseous moles of the product is equal to the gaseous moles of the reactant.

Hence, the no work done by the system.

D. $CaCO_3 (s) \rightarrow CaO (s) + CO_2(g) ,\Delta H = 110.1 kJ/mol$

The increase in the volume of the system as gaseous moles of product is more than gaseous moles of reactant

Hence, the work done by the system.

E. $4 NH_3(g) + 5 O_2(g) \rightarrow 4 NO (g) + 6 H_2O(l) ,\Delta H = -906 kJ/mol$

The decrease in the volume of the system as gaseous moles of product is less than gaseous moles of reactant.

Hence, the work done by the surroundings.

Learn more about the system and surroundings here:

link

Answer from: Yehh7630

A and D: system does work

B and E: surroundings do work

C: No work done by system or surroundings

Explanation:

A system does work on the surroundings if the volume increases, because it is pushing back against the atmosphere.

The volumes of liquids don't change much during a reaction. However, if a gas is absorbed or released at constant pressure, the volume changes significantly.

Thus, we can ignore the solids and liquids and consider only whether there is a change in the number of moles of gas.

The formula for the work done is

w = -pΔV = -ΔnRT where

Δn = n₂ - n₁ = moles of product gases - moles of reacting gases

If Δn is +, w is -, and the system does work on the surroundings.

If Δn is -, w is +, and the surroundings do work on the system.

A. H₂O(ℓ) ⟶ H₂O(g); ΔH = 44.0 kJ/mol

Δn = 1 - 0 = 1

w = -1RT = -RT, so the system does work.

B. 2NO(g) + O₂(g) ⟶ 2NO₂(g); ΔH = -114.1 kJ/mol

Δn = 2 - 3 = -1

w = 1RT = RT, so the surroundings do work.

C. Cl(g) + O₃ (g) ⟶ ClO(g) + O₂(g); ΔH = -163 kJ/mol

Δn = 2 - 2 = 0

w = 0, so the system does no work.

D. CaCO₃(s) ⟶ CaO s) + CO₂(g); ΔH = 110.1 kJ/mol

Δn = 1 - 0 = 1

w = -1RT = -RT, so the system does work.

E. 4NH₃(g) + 5O₂(g) ⟶ 4NO(g) + 6H₂O(l); ΔH = -906 kJ/mol

Δn = 4 - 9 = -5

w = 5RT, so the surroundings do work.

Answer from: Quest

a saline water is fed to a reverse osmosis unit at flow rate o rotameter connected to reject (retentate) line reads 8 installed in reject and permeate line reads 15.2 milli siemens/cm and 300 micro siemens/com respectively.

Answer from: Quest

Which of the following did you identify correctly

Another question on Chemistry

Chemistry, 21.06.2019 21:30

Calculate the h3o+ concentration in a solution of acetic acid if the concentration of molecular acetic acid present at equilibrium is 9.97x10^-3 m and k for the dissociation is 1.86x10^-5. ch3cooh(aq)+h2o(> h3o^+(aq)+ch3coo^-(aq)

Answers: 2

Answer

Chemistry, 21.06.2019 22:30

Often on a topographic map, every fifth contour line is darkened. what is this line called? a. key b.slope c.benchmark d. index contour

Answers: 1

Answer

Chemistry, 22.06.2019 06:00

24. a sports ball is inflated to an internal pressure of 1.85 atm at room temperature (25 °c). if the ball is then played with outside where the temperature is 7.5 °c, what will be the new pressure of the ball? assume the ball does not change in volume nor does any air leak from the ball a) 0.555 atm b) 1.74 atm c) 1.85 atm d) 1.97 atm

Answers: 2

Answer

Chemistry, 22.06.2019 12:20

Adeuteron, 21h, is the nucleus of a hydrogen isotope and consists of one proton and one neutron. the plasma of deuterons in a nuclear fusion reactor must be heated to about 3.02×108 k . what is the rms speed of the deuterons? express your answer using two significant figures.

Answers: 1

Answer

You know the right answer?

Select whether there is no work done by the system, work done by the system or work done by the surr...

Questions