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pochemuha
3 years ago
7

vaporized at 100°C and 1 atmosphere pressure. Assuming ideal gas 1 g mole of water is behavior calculate the work done and compa

re this with the latent heat (40.57 kJ/mole). Why is the heat so much larger than the work?
Chemistry
1 answer:
Vesna [10]3 years ago
7 0

Answer:

q = 40.57 kJ; w = -3.10 kJ; strong H-bonds must be broken.

Explanation:

1. Heat absorbed

q = nΔH = 1 mol × (40.57 kJ/1 mol) = 40.57 kJ

2. Change in volume

V(water) = 0.018 L

pV = nRT

1 atm × V = 1 mol × 0.082 06 L·atm·K⁻¹mol⁻¹ × 373.15 K

V = 30.62 L

ΔV = V(steam) - V(water) = 30.62 L - 0.018 L = 30.60 L

3. Work done

w = -pΔV = - 1 atm × 30.60 L = -30.60 L·atm

w = -30.60 L·atm × (101.325 J/1 L·atm) = -3100 J = -3.10 kJ

4. Why the difference?

Every gas does 3.10 kJ of work when it expands at 100 °C and 1 atm.

The difference is in the heat of vaporization. Water molecules are strongly hydrogen bonded to each other, so it takes a large amount of energy to convert water from the liquid phase to the vapour phase.

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The bond dissociation energy of the Cl - Cl bond is -958 kJ mol^-1.

<h3>What is the dissociation enthalpy?</h3>

Given that;

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ΔHfO of HCL(g) = -92kJ mol^-1

Bond dissociation enthalpy of the Cl-Cl bond = x

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2 years ago
How do trends in the periodic table help predict the properties of an element
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3 years ago
There are two steps in the usual industrial preparation of acrylic acid, the immediate precursor of several useful plastics. In
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Answer:

The net change in enthalpy for the formation of one mole of acrylic acid from calcium carbide, water and carbon dioxide is 523.2 kJ.

Explanation:

Step 1:

CaC_2(s) + 2H_2O(g)\rightarrow C_2H_2(g) + Ca(OH)_2(s),\Delta H_1=414.0 kJ...[1]

Step 2 :

6C_2H_2(g) + 3CO_2(g) + 4H_2O(g)\rightarrow 5CH_2CHCO_2H(g) \Delta H_2=132.0kJ..[2]

Adding 6 × [1] and [2]:

6CaC_2(s) + 12H_2O(g)\rightarrow 6C_2H_2(g) + 6Ca(OH)_2(s)

6C_2H_2(g)+3CO_2(g)+16H_2O(g)\rightarrow 5CH_2CHCO_2H(g)

we get :

6CaC_2(s) + 8H_2O(g)+3CO_2(g)\rightarrow 5CH_2CHCO_2H(g)+ 6Ca(OH)_2(s),\Delta H'=?

\Delta H'=6\times \Delta H_1+\Delta H_2

\Delta H'=6\times 414.0 kJ+132.0kJ

\Delta H'=2,626 kJ

Energy released on formation of 5 moles of acrylic acid = 2,626 kJ

Energy released on formation of 1 mole of acrylic acid:

\frac{ 2,626 kJ}{5 } = 523.2 kJ

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