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

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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:

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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.

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