Chemical Thermodynamics:

The first lawHess’s law

Standard enthalpies of formation

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First Law of Thermodynamics

The energy cannot be created or destroyed. Energy can, however, be converted from one form to another or transferred from a system to the surroundings:

Q = ΔU + W,

where Q – an amount of heat;

ΔU – changes in internal energy;

W – a work resisting external forces:

W = PΔV

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An 80.0-gram sample of a gas was heated from 25 °C to 225 °C. During this process, 346 J of work was

done by the system and its internal energy increased by 9055 J. What is the specific heat of the gas?

Solution

The first principle calculations

𝛥𝑈 = 𝑄 −𝑊

𝑄 = 𝛥𝑈 +𝑊

𝑄 = 𝑐𝑚𝛥𝑇

𝛥𝑈 +𝑊 = 𝑐𝑚𝛥𝑇

𝑐 =𝛥𝑈 +𝑊

𝑚𝛥𝑇

𝑐 =9055 + 346

80 ∗ (225 − 25)= 0.588

𝐽

𝑔 ∗ °C

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Thermodynamic State Functions: Thermodynamic properties that are dependent on the state of the system only. (Example: DE and DH)

Other variables will be dependent on pathway (Example: q and w). These are NOT state functions. The pathway from one state to the other must be defined.

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Enthalpy – heat content

If a system is exchanged with surroundings by heat its’ internal energy and enthalpy are changed:

Q > 0 (heat is liberated) ΔU < 0, ΔH < 0;

Q < 0 (heat is absorbed) ΔU > 0, ΔH > 0.

The value of ΔH is corresponding to the amount of the heat Q which is absorbed or evolved in any chemical reaction:

Q = – ΔH

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Standard Enthalpy of Formation

The standard enthalpy of formation ΔHf is the enthalpy change when 1 mol of a pure substance is formed from its elements.

Each element must be in the physical and chemical form which is most stable at normal atmospheric pressure (101325 Pa or 1 atm) and a specified temperature (usually 298 K or 25 °C).

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If ΔHf[H2O(l)] = –285.8 kJ· mol–1 the thermochemical equation is:

H2(g) + ½O2(g) → H2O(l) + 285.8

ΔHm = –285.8 kJ·mol–1

The standard enthalpy of formation for a simple substance in its most stable state must be

zero.

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Hess’ Law: DH for a process involving the transformation of reactants into products is not dependent on pathway. Therefore, we can pick any pathway to calculate DH for a reaction.

When calculating DH for a chemical reaction as a single step, we can use combinations of reactions as “pathways” to determine DH for our “single step” reaction.

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Once can always reverse the direction of a reaction when making a combined reaction. When you do this, the sign of DH changes.

N2(g) + 2O2(g) 2NO2(g) DH = 68 kJ

2NO2(g) N2(g) + 2O2(g) DH = -68 kJ

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Our reaction of interest is:N2(g) + 2O2(g) 2NO2(g) DH = 68 kJ

• This reaction can also be carried out in twosteps:

N2 (g) + O2 (g) 2NO(g) DH = 180 kJ2NO (g) + O2 (g) 2NO2(g) DH = -112 kJ

If we take the previous two reactions and add them, we get the original reaction of interest:

N2 (g) + O2 (g) 2NO(g) DH = 180 kJ2NO (g) + O2 (g) 2NO2(g) DH = -112 kJ

N2 (g) + 2O2 (g) 2NO2(g) DH = 68 kJ

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To calculate ΔH we add all the ΔHf values for the products, multiplying each by the appropriate coefficient.

Than we add all the ΔHf values for the reactants again multiplying by appropriate coefficients. This can be summarized by the equation

ΔH = ∑ iΔHf (products) – ∑ jΔHf (reactants)

the symbol Σ means “the sum of”,

i,j – coefficients in the reaction.

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Write the balanced incomplete combustion reaction for ethane (C2H6), producing CO and watervapour. What is the enthalpy of the reaction (ΔH°rxn), in kJ, for the process?

2C2H6(g) + 5O2 (g)= 4CO (g)+ 6H2O(g)

ΔH°rxn = 4ΔH° CO + 6ΔH° H2

O − 2ΔH° C2H

6

− 5ΔH° O2

ΔH°rxn = 4ΔH° CO + 6ΔH° H2

O − 2ΔH° C2H

6

− 5ΔH° O2

ΔH°rxn = 4 ∗ −110.5 + 6 ∗ −241.8 − 2 ∗ −84 − 5 ∗ 0 = −𝟏𝟕𝟐𝟒. 𝟖 𝒌𝑱

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