Enthalpies for Different Types of Reactions: Enthalpy, Videos, Examples

How do you think different reactions take place? Do you think it is always spontaneous? No, right? There is something that we know as the enthalpy of a reaction. Are you aware of this term? Well, in the study of various chemical reactions, we must know about the enthalpy and standard enthalpy of the same. Why is it important? Well, let’s find out in the chapter below. We will also cover the enthalpies of different types of reactions.

Why is Enthalpy Important?

The concept of enthalpy is vital for temperature and pressure that any chemical reaction requires. We also need to know the enthalpy to calculate the total amount of heating and cooling that a commercial production of any substance requires.

For example, in the mass production of any compound such as ammonia, calcium carbonate, oxygen etc. we need to know the enthalpy in the reaction. So, how do we calculate the change in enthalpy for any reaction?

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Enthalpy change

Enthalpy change is the standard enthalpy of formation. We have determined it for a vast number of substances. In any given chemical reaction, the reactants undergo chemical changes and combine to form different products.

For any such reaction, we represent the enthalpy change as Δ_rH. We term it as the reaction enthalpy. We can calculate the reaction enthalpy by subtracting the sum of enthalpies of all the reactants from that of the products. Mathematically,

Δ_tH = Sum of enthalpies of the product – Sum of the enthalpies of the reactants

This is a brief outline of the concept of enthalpy. However, our main aim in this chapter is to earn about the enthalpies of different reaction. Let us now look at these.

Types of Reactions and Their Enthalpies

Below, we will look at the various types of reactions and their associated enthalpies.

Enthalpy of Formation

We can define it as “The quantity of heat evolved or absorbed when one mole of the compound forms from its elements”. We can express it as $\Delta Hf$ , e.g.

$\dfrac{1}{2} H_2 (g) + \dfrac{1}{2}I_2 (s) \to HI(g) \hspace{5mm} \Delta H_f = +6.2 kcal$

$H_2 (g) + \dfrac{1}{2} O_2(g) \to H_2 O (l) \hspace{5mm} \Delta H_f = -68 kcal$

$\dfrac{1}{2} N_2 (g) + \dfrac{3}{2} H_2 (g) \to NH_3 (g) \hspace{5mm} \Delta H_f = -11 kcal$

It is important to remember that we should balance the thermochemical equation in such a way that it represents the formation of a single mole of the substance only. Standard heat of formation is the value of the heat of formation at 298 K and 1 atm pressure. We take it as zero in case of the free state of elements.

Enthalpy

Enthalpy of Combustion

We can define it as, “The quantity of heat evolved when one mole of the substance completely oxidises”. Eg: $(Delta H = -ve)$ always exothermic

$\underset{diamond}{C} + O_2 \to CO_2 + 95.5 kcal$

$CH_4 (g) + 2O_2 (g) \to CO_2 (g) + 2H_2 O (l) \hspace{2mm} \Delta H = -213 kcal$

$C(diamond) + O_2 (g) \to CO_2 (g) \hspace{10mm} \Delta H = -95.5 kcal$

$C(graphite) + O_2 (g) \to CO_2 (g) \hspace{10mm} \Delta H = -94.0 kcal$

$H_2 (g) + \dfrac{1}{2} O_2 (g) \to H_2O (l) \hspace{10mm} \Delta H = -68.4 kcal$

The heat of combustion finds its use in the calculation of the heat of formation which looks difficult in certain cases, like in the calculation of the calorific value of fuels. The heat of combustion is also useful in elucidating or explaining the structure of various organic compounds.

Enthalpy of Solution

We may define it as, “The quantity of heat evolved or absorbed when one mole of a solute totally dissolves in a large excess of water, so that further dilution of solution does not produce any heat change”. E.g.

$KCl(s) + nH_2O \to KCl(aq) \Delta H = +4.40 kcal$

$HCl (g) + nH_2O \to HCl (aq) \Delta H = -39.2 kcal$

Enthalpy of Neutralisation

It is nothing but, “the quantity of heat evolved when one equivalent (or equivalent mass) of an acid is completely neutralised by one equivalent (or equivalent If mass) of a base in dilute solution”. E.g.

$HNO_3 (aq) + NaOH (aq) \to NaNO_3 (aq) + H_2O (l) \Delta H = -13.69 kcal$

$HCl (aq) + NaOH (aq) \to NaCl(aq) + H_2O (l) \Delta H = -13.68 kcal$

We take the heat of neutralisation of strong acid and a strong base as 13.7 kcal. or 57kJ. We can use the electrolytic dissociation theory to explain that this heat of neutralisation is merely the heat of formation of water from $H^+$ of an acid and $OH^-$ of a base.

The heat of neutralisation for some weak acids or weak bases is a little lesser than 13.7 kcal. This is because some energy is used up in dissociating the weak electrolyte. We can calculate the dissociation energy of the weak acid or base by calculating the difference in the values.

$NH_4OH(w. base) + HCl(Strong acid) \leftrightharpoons NH_4Cl + H_2O \Delta H = -12.3 kcal$

$CH_3COOH (weak acid (aq)) + NH_4OH(weak base(aq)) \leftrightharpoons CH_3COOHNH_4 (aq) + H_2O(l) \Delta H = -11.9 kcal$

Here, 1.8 kcal of heat is used in the dissociation of both the weak electrolytes. We can measure the heat of neutralisation in the lab by using pothythene or polystyrene bottles.

Enthalpy of Dissociation

We can define this term as, “The quantity of heat absorbed when one mole of a substance completely dissociates into its ions”. E.g.

$H_2O(l) \to H^+ + oh^- \Delta H = 13.7 kcal$

Enthalpy of Dilution

We can define it as, “The quantity of heat evolved or absorbed when a solution containing one mole of a solution dilutes from one concentration to another”.

$KCl (S) + 20H_2O \to KCl(20 H_2O) \Delta H_1 = +3.8 kcal$

$KCl(s) + 200 H_2O \to KCl(200 H_2O) \Delta H_2 + 4.44 kcal$

$\therefore$ Heat of dilution = $\Delta H_2- \Delta H_1 = 4.44- 3.80 = 0.64 kcal$

Enthalpy of Precipitation

We can define it as, “The quantity of heat given out in the precipitation of one mole of a sparingly soluble substance on mixing dilute solutions of suitable electrolytes”. E.g.

$Ba^{2+} (aq) + SO^{2-}_4 (aq) \to BaSO_4 (s), \Delta H = -4.66 kcal$

Enthalpy of Hydration

We may define it as, “The quantity of heat evolved or absorbed? when one mole of an anhydrous or a partially hydrated salt is combined with the required number of moles of water to form a specific hydrated substance”. E.g.

$CuSO_4 (s) + 5H_2O (l) \to CuSO_4. 5H_2O (s) \Delta H = -18.7 kcal$

Enthalpy of Fusion

We can define it as, “It is the change in enthalpy during conversion of 1 mole of a substance from solid to liquid state at its melting point (mostly endothermic)”.

$H_2O (solid) \overset{\Delta}{\underset{m.p.} \rightarrow} H_2O(liquid)- 1.44 kcal$