Chemistry : Thermodynamics

Thermodynamics

Specific heat capacity (c) is the amount of heat required to raise 1g of the substance by 1K. Hence its unit is Jg^-1K^-1. Note that the unit used is different from that of in physics (use kg instead of g). Water has specific heat capacity 4.18 Jg^-1K^-1. So 4.18 J of energy raises 1g of water by 1 K.

Heat capacity (C) is the amount of heat required to raise the substance by 1K, in JK^-1.

Mathematically, E (energy change of a substance) = mcΔT = CΔT (C = mc).

Law of conservation of energy states that energy can’t be created or destroyed, they can only transform into different forms.

We are not interested in the absolute energy of a system, but we can calculate the change in internal energy (E) of a system. ΔE = E_final - E_initial = q + w, where q is the heat released by the system and w is the work done against the pressure (related to PV=nRT). However, when the system is kept at constant pressure w = 0, then we can define enthalpy change ΔH is the heat released or taken in during any change in a system of constant pressure. ΔH = H_product – H_reactants.

In an exothermic reaction, heat is released to the surroundings, so the internal energy decreases (energy transformed from internal energy in substances to heat in surroundings), ΔH<0.

Endothermic reactions means that heat is absorbed from the surroundings so its internal energy increases, ΔH>0.

Physical change may lead to enthalpy change as well, like the change of state from H₂O(l) to H₂O(g) is endothermic reactions (latent heat is absorbed).

Enthalpy diagram shows the relative level of enthalpy of the substances, with enthalpy + unit as the y-axis, there’s no x-axis unless it’s a reaction coordinate. Note that all enthalpy change with +/- sign, and state of substances (s/l/g/aq) must be given.

Thermochemical equation shows more data than normal chemical equations.

1)       It shows the state and enthalpy change of reactions, the unit is kJ mol^-1, e.g. CH₄(g)+2O₂(g)→CO₂(g)+2H₂O(l) ΔH = -890 kJ mol^-1

2)       The enthalpy change depends on # of moles, e.g., 2CH₄(g)+4O₂(g)→2CO₂(g)+4H₂O(l) ΔH = -1780 kJ mol^-1 which is double of the above equation.

3)       ΔH of reversed reaction direction = -ΔH of original reaction. E.g., CO₂(g)+2H₂O(l)→CH₄(g)+2O₂(g) ΔH = +890 kJ mol^-1 (Note: you must write a “+”)

We can explain the difference in enthalpy by the breakage/formation of bonds. Breaking the bonds are endothermic while forming bonds are exothermic. If the combined result is exothermic, the overall enthalpy change is less than zero. For example we can look the combustion of methane as breaking CH₄ and H₂O into C, H, O and reform into CO₂ and H₂O.

The standard condition includes (1): 298K, (2) 1atm, (3) substances at 298K, 1atm and in the most stable physical state (include the allotrope form, e.g. C(graphite) is more stable then C(diamond) so C(graphite) is chosen as the substance representing carbon, (4) solutions are at 1M. Such reaction we can call as standard condition for measuring ΔH, denoted by ΔH^o.

1)       Standard enthalpy change of reaction ΔH^o_r, is the enthalpy change when molar quantities of reactants as started in the specified equation react together under standard conditions. e.g., CO₂(g)+2H₂O(l)→CH₄(g)+2O₂(g) ΔH = +890 kJ mol^-1. (This is combustion, the term reaction usually refers to reactions other than combustion, like neutralization, etc.)

2)       Standard enthalpy change of formation ΔH^o_f of a substance is the enthalpy change when one mole of the substance is forms from its elements in their standard states. Standard states refer to H₂, O₂, Ar, Au, etc. (not H, O), e.g., consider 1/2 N₂(g) + O₂(g)→NO₂(g),  ΔH^o_f[NO₂(g)]=+39.9 kJ mol^-1. Note that it can be endothermic like the formation of ethyne. The higher ΔH during the formation usually implies that the product is more unstable. For example, H₂O is more stable then H₂ and O₂ while C₂H₂ (ethyne) is less stable then C and H₂.

3)       Standard enthalpy change of combustion ΔH^o_c of a substance is the enthalpy change when one mole of the substance is completely brunt in O₂ in standard condition. For example, the production of CO is not complete combustion while CO₂ is. Again CH₄(g)+2O₂(g)→CO₂(g)+2H₂O(l) ΔH = -890 kJ mol^-1 is an example of enthalpy change of combustion.

4)       Standard enthalpy change of neutralization ΔH^o_n is the enthalpy change when acid and alkali react to give 1 mole of water in standard conditions. ΔH^o_n between a strong monobasic acid(HCl) and strong alkali is about -57.1 kJ mol^-1 while ΔH^o_n between a strong dibasic acid (H₂SO₄) and a strong alkali is about -64.8 kJ mol^-1. It must be less than the above value if weak acid/alkali is used since energy is used to dissociate the acid/alkali before dissociation.

5)       Standard enthalpy change of solution ΔH^o_s is the enthalpy change when 1 mole of substances dissolves in infinite amount of water (enough solvent that further dilution has no effect) under standard conditions. It can be exothermic (e.g. dissolution of NaCl) or endothermic (LiCl)

Measurement

1)       ΔH^o_n is measured in a calorimeter: a vacuum flask containing a stirrer, thermometer and the reactant solutions with stopper on the top. We measure the energy released by measuring the temperature rise. In laboratory we can use two expended polystyrene cups with a lid instead of calorimeter. At the instant that it reacts, the temperature may not be accurate. By plotting temperature T – time t graph (red) , we extend the line (blue) to find the max. temperature rise.

2)       ΔH^o_c is measured in similar settings. The weight of the spirit lamp containing the substance is measured before and after ignition. We use a known mass of water in the cup. By comparing energy absorbed by the water and the mass of substances combusted we can calculate the enthalpy change.

Note that draught screen can be used to reduce heat loss to surroundings.

Flame calorimeter is an improved setup for more accurate result.

Source of error:

-              Heat loss to surroundings (universal)

-              Heat capacities of thermometer/cup/stirrer are neglected

-              Density and heat capacity of solution aren’t exactly equal to that of water (if it’s assumed to be similar as water.

-              Thermometer not sensitive enough (this reason is not preferred)

Hess’ Law states that the enthalpy change of a reaction depends on the initial and final state of the reaction, independent of the route by which reaction may occur.

Analysis

By Born-Haber cycle: Route 1 (red) vs Route 2 (blue) When we try to find the enthalpy change from reactants to intermediate product, we can’t put the reactants together since it reacts directly into products. For example the ΔH^o_r between C and CO can’t be determined by preparing C and O₂ because they react directly to gives CO₂. We have ΔH^o_r = ΔH^o_c[C(graphite)] - ΔH^o_c[CO(g)].

Similar cases always occur: we can’t directly obtain the enthalpy change by experiment because it reacts spontaneously to give other products. For example you can’t obtain the enthalpy change of ethene by combining C and H because they readily react with oxygen to give CO₂ and HO.

Another example: the standard enthalpy change of formation of NaOH – direct reaction only gives NaOH(aq), NOT NaOH(s).

We prepare excess hydrogen (1/2 H₂) as dummy.

Route 1: Na(s) + 1/2 O₂(g) + H₂(g) → Na(s) + H₂O(l) → NaOH(aq) + 1/2 H₂(g)

ΔH = ΔH^o_c[H₂(g)] + ΔH^o_r

Route 2: Na(s) + 1/2 O₂(g) + H₂(g) → NaOH(s) + 1/2 H₂(g) → NaOH(aq) + 1/2 H₂(g)

ΔH = ΔH^o_f[NaOH(s)] +ΔH^o_s[NaOH(s)]

Therefore ΔH^o_f[NaOH(s)] = ΔH^o_c[H₂(g)] + ΔH^o_r - ΔH^o_s[NaOH(s)]

Sometimes we need other dummies like HCl in determining ΔH^o_f[MgCO₃(s)].

Enthalpy diagram is another way to show the Born-Haber cycle as shown in the figure.

The third method is by algebraic addition or subtraction to obtain the required enthalpy change.

e.g. ΔH^o_f[ZnS(s)]

(1): Zn(s) + 1/2 O₂(g) → ZnO(s) ΔH = -348 kJ mol^-1

(2): ZnO(s) + SO₂(g) → ZnS(s) + 3/2O₂(g) ΔH = +441 kJ mol^-1

(3): S(s) + O₂(g) → SO₂(g) ΔH = -297 kJ mol^-1

By crossing out the dummies we have

Zn(s) + S(s) → ZnS(s) ΔH = (-348+441-297) = -204 kJ mol^-1

The advantage is that we can directly find out the enthalpy change without clearly figuring the route of reaction. Take this 1991 HKALE question as example:

Find ΔH^o_f[NaCl(s)] given the following thermodynamic equation:

(1): NaOH(aq) + HCl(aq) → NaCl(aq) + H₂O(l) ΔH = -57.3 kJ mol^-1

(2): H₂(g) + 1/2 O₂(g) → H₂O(l) ΔH = -258.9 kJ mol^-1

(3): 1/2 H₂(g) + 1/2 Cl₂ (g) → HCl(g) ΔH = -92.3 kJ mol^-1

(4): HCl(g) → HCl(aq) ΔH = -71.9 kJ mol^-1

(5): Na(s) + 1/2 O₂ + 1/2 H₂ + aq → NaOH(aq) ΔH = -425.6 kJ mol^-1

(6): NaCl(s) → NaCl(aq) ΔH = +3.9 kJ mol^-1

We reverse (6) to get (6)’ NaCl(aq) → NaCl(s) ΔH = -3.9 kJ mol^-1 and reverse (2) to get (2)’: H₂O(l)  → H₂(g) + 1/2 O₂(g) ΔH = +285.9 kJ mol^-1

By adding up (1), (2)’, (3), (4), (5), (6)’, we have Na(s) + 1/2 O₂(g) + 1/2 H₂(g) → NaOH(s) where  ΔH = -57.3+285.9-92.3-71.9-425.6-3.9 = -365.1 kJ mol^-1

Two shortcuts:

1)       By considering reactant = constituent elements, dummy = oxygen, we have

ΔH^o_f[compound] = ΣΔH^o_c[constituent elements] -ΔH^o_c[compound]

For example, ΔH^o_f[C₂H₅OH(l)] = 2ΔH^o_c[C(graphite)] + 3ΔH^o_c[H₂(g)] - ΣΔH^o_c[C₂H₅OH(l)]

Note that ΔH^o_c[O₂(g)] does not exist since oxygen do not react with itself.

2)       More generally we have ΔH^o of a certain reaction = ΣΔH^o_f[products] - ΣΔH^o_f[reactants]

For example, ΔH^o_n = ΣΔH^o_f[salt + water] -ΔH^o_f[acid + alkali] for neutralization.