ENTHALPY AND ENTHALPY CHANGE

Enthalpy (H) is a property of a substance that can be used to obtain the evolved or absorbed in a chemical reaction. When a chemical reaction occurs, the total enthalpy (that is the total heat content) of the system changes and this enthalpy is denoted as ∆H. the change in enthalpy of a reaction at a given temperature and pressure (called enthalpy of reaction) is obtained by subtracting the total enthalpy of the reactant from the total enthalpy of the products.

∆H reaction = Ɛn HP – Ɛm Hr ------------------------ 1

Where Ɛ represents summation.

N and m are the coefficients of the substances in the products and reactants respectively.

Hp = enthalpies of the products

Hr = enthalpies of reactants

The values of enthalpies are normally given as standard enthalpies, ∆H^Ѳ , which are for reactants and products in their normal states at 298k and 1atm pressure (called STP).

The change in enthalpy of a reaction if often given with the chemical equation, that is, as a thermochemical equation, for example:

CH4_(g) + 2O₂ ----------- CO₂ + 2H₂0; H = -890KJ/mol

In writing H = -890kj/mol, we mean that 1 mole of methane reacts with 2 moles of oxygen to yield 1 mole of carbon dioxide and 2 moles of water. Also, it is important to give the state of each substance in chemical equations because, in general, change in H will be different if the state of any reactant or product changes. Let us consider the chemical equations:

CH4_(g) + 2O2_(g) ------------ CO2(g)  +  2H₂O_(l); H =-890kj/mol………….i

 CH4_(g) + 2O2_(g) ------------ CO2(g)  +  2H₂O_(g); H =-802kj/mol………….ii

Comparing the enthalpies of the reactions (i) and (ii), one would notice that less heat is evolved for chemical reaction (ii) than for chemical reaction (i) (802kj/mol versus 890kj/mol). In the chemical reaction (ii), water is obtained as the vapour, i.e. H₂O_(g), instead of the liquid, H₂O_(l) as in chemical reaction (i). the lower enthalpy of reaction is therefore due to the fact that some heat is used to vapourize liquid water, and consequently less heat is released to surroundings.

WORKED EXAMPLE

Question1: How much heat is evolved when 500kg of ammonia is produced according to the following equation:

N_2(g) + 3H_2(g) --------- 2NH_3(g); H = -91.8KJ.   [N = 14.0, H = 1.0 g/mol].

SOLUTION:

n_NH3     =   molar given/ molar mass of ammonia

molar mass of NH₃ = 14 + 3(1.0) = 17 g/mol

mass given = 500 x 10³ g

therefore,  n_NH3 = 500 x 10³ g /17 g/mol

                              =  2.94 x 10⁴ mol.

From the equation given,

2 moles of NH₃ gave -91.8kj

Therefore,  2.94 x 10⁴ moles of NH₃ will give

(-91.8 kj / 2) X (2.94 x 10⁴ moles)

= 1,35 x 10⁶ kj.