How to calculate delta H (∆H) of a reaction?

The Greek letter delta (∆) denotes change, while H stands for enthalpy. Thus, ∆H (Delta H) represents the enthalpy change for a chemical reaction.

Enthalpy refers to the total energy content of a system.

Therefore, ∆H drives the fundamentals of all chemical reactions. It helps in determining the total amount of heat exchanged between the system and its surroundings.

What does ∆H (Delta H) tell us?

Whenever a chemical reaction takes place, old reactant bonds break while new product bonds are formed.

Bond breaking requires energy input (+ q), while bond formation releases a certain amount of energy (-q) into the surroundings.

Based upon the above concept, chemical reactions are majorly categorized into two types, i.e., endothermic and exothermic.

Endothermic reaction: The amount of heat supplied for the reaction to take place is greater than the energy released during the reaction, resulting in a positive enthalpy change (∆H > 0).

Exothermic reaction:  The amount of energy released into the surroundings is greater than the heat supplied to the system. This leads to a negative enthalpy change (∆H < 0).

Thus, ∆H helps in predicting whether a chemical reaction is exothermic or endothermic in nature.

Now let us find out, with the help of some examples given below, how to calculate ∆H under different situations.

How to calculate ∆H (Delta H) for a reaction? – Examples

For a general chemical reaction, reactants → products, the enthalpy change is calculated as follows:

Where,

• ∆H (reaction)= Enthalpy change for the reaction
• Σ∆H_f^° (products) = Sum of the standard enthalpies of formation of products
• Σ∆H_f^° (reactants) = Sum of the standard enthalpies of formation of reactants

∆H_f^° stands for enthalpy of formation of a compound, i.e., the amount of energy released when 1 mole of a compound is formed from its constituent elements under gaseous conditions (273.15 K temperature and 1 atm pressure).

The units of enthalpy change (∆H) are J/mol or kJ/mol.

Let’s see an example of finding ∆H: Calculate Delta H for the reaction given below:

4 NH_{3 (g)} + 7 O_{2 (g)} → 4 NO_{2 (g)} + 6 H₂O _(l)

Given that

• ∆H_f^° (NH₃) = -46.1 kJ/mol
• ∆H_f^° (NO₂) = 33.2 kJ/mol
• ∆H_f^° (H₂O) = -286 kJ/mol

Solution:

Modifying equation (i) w.r.t the reaction shown above gives us:

∆H (reaction) = [(6 x ∆H_f° (H₂O) + (4 x ∆H_f^° (NO₂)] – [4 x ∆H_f° (NH₃) + 7 x ∆H_f° (O₂)]

Important: Do not forget to multiply ∆H_f^° for each of the reactants and products with their respective number of moles as per the balanced chemical equation.

Also, ∆H_f° (O₂) = 0, as enthalpy of formation is only valid for compounds, while oxygen is a molecule and not a compound.

Substituting all the known values to find unknown ∆H (reaction).

∆H (reaction) = [6(-286) + 7(33.2)] – [4(-46.1)]

∆H (reaction) = [-1716 + 232.4] + 184.4

∆H (reaction) = – 1483.6 + 184.4

∴ ∆H (reaction) = -1299.2 J/mol = – 1.2992 = 1.3 kJ/mol

Result:  The enthalpy change (∆H) for the given chemical reaction is -1.3 kJ/mol. A negative value of ∆H (Delta H) implies that the reaction is exothermic in nature.

Another example is– Calculate ∆H (Delta H) for the reaction given below:

B₂O_{3 (s)} + 3 H₂O _(g) →3 O_{2 (g)} + B₂H_{6 (g)}

Given that

• ∆H_f^° (B₂O₃) = -1263.6 kJ/mol
• ∆H_f^° (H₂O) = -286 kJ/mol
• ∆H_f^° (B₂H₆) = +31 kJ/mol

Is this reaction exothermic or endothermic?

Solution:

Modifying equation (i) w.r.t the reaction shown above:

∆H (reaction) = [∆H_f^° (B₂H₆)] – [∆H_f° (B₂O₃) + 3 x ∆H_f° (H₂O)]

Substituting all the known values:

∆H (reaction) = 31 – [-1263.6 + 3(-286)]

∆H (reaction) = 31 – (-2121.6)

∴ ∆H (reaction) = 2152.6 J/mol = 2.15 kJ/mol

Result:  The enthalpy change (∆H) for the given chemical reaction is + 2.15 kJ/mol. A positive value of ∆H implies that the reaction is endothermic in nature.

Formulae to calculate ∆H (Delta H) in a thermodynamic system  

Formula # 1

 The formula q = n∆H can be applied to find the heat of a reaction (q) if the number of moles (n) and enthalpy change (∆H) are given.

Conversely, it can be rearranged to make Delta H the subject of the formula as follows:

q = n ∆H

Where,

• ∆H = Enthalpy change (Units: J/mol or kJ/mol)
• q = Heat exchanged between the system and its surroundings (Units: J or kJ)
• n = Number of moles of reactant (Units: mol)

For example, 361.2 kJ of energy is absorbed when 2 moles of nitrogen react with an excess of oxygen to produce nitrogen monoxide as the only product.

N_{2 (g)} + O_{2 (g)} → 2 NO _(g)

Calculate ∆H for the reaction.

Solution:

As per the question statement,

q = + 361.2 kJ

n = 2 moles

∆H =?

Applying the formula and substituting all the known values:

⇒ ∆H = q/n

∆H = 361.2/2

∴ ∆H = 180.6 kJ/mol

Result: The enthalpy change for the reaction is +180.6 kJ/mol. Heat is absorbed during the reaction, and a positive ∆H value implies that it is an endothermic reaction.  

Formula # 2

Another important formula in thermodynamic systems is:

∆G = ∆H – T∆S…..Equation (iii)

Where,

• ∆G = Change in Gibbs free energy of the system (Units: J/mol or kJ/mol)
• ∆H = Enthalpy change (Units: J/mol or kJ/mol)
• T = Absolute temperature (Unit: K)
• ∆S = Entropy change (Units: J/K.mol or kJ/K.mol)

Equation (iii) helps in predicting the spontaneity and direction of a chemical reaction, in addition to finding out its endothermic or exothermic nature.

It can be rearranged to make ∆H the subject of the formula:

∆H = ∆G + T∆S…..Equation (iv)

Now let’s find ∆H using the above equation- Calculate ∆H for the reaction shown below that proceeds at T= 298 K, given that its Gibbs free energy change (∆G) is + 578.4 kJ/mol and entropy change (∆S) is 7.5 x 10^-3 kJ/K.mol at room temperature. 

PCl_{3 (g)} + 3 HCl _(g) → 3 Cl_{2 (g)} + PH_{3 (g)}

Solution:

As per the question statement;

∆G = 578.4 kJ/mol

T = 298 K

∆S = 7.5 x 10^-3  kJ/K.mol

∆ H (Delta H) =?

Applying the formula:

⇒ ∆H = ∆G + T∆S

∆H = 578.4 + 298(7.5 x 10^-3)

∆H = 578.4 + 2.235

∴ ∆H = 580.6 kJ/mol

Result: The enthalpy change (∆H) for the reaction is + 580.6 kJ/mol. Thus, the reaction is endothermic in nature.

Formula # 3

The energy input or expelled by a system is related to the specific heat capacity of the substance by the formula given below:

q = mc∆T

q = ∆H if n =1, thus the above equation is transformed into:

∆H = mc∆T……Equation (v)

Where,

• ∆H (Delta H) = Enthalpy change (Units: J/mol or kJ/mol)
• m = mass of the substance (Unit: kg)
• c = specific heat capacity (Units: J.kg^-1°C^-1)
• ∆T = change in temperature (Units: °C)

For example, 36 grams of water is heated such that its temperature changes from 25°C to 67°C. Calculate enthalpy change (∆H) if the specific heat capacity of water is 4200 J/kg°C.

Solution:

Given in the question statement:

m= 36 g = 0.036 kg

c = 4200 J/kg°C

∆T = 67 – 25 = 42°C

Applying the formula:

⇒ ∆H = mc∆T

Substituting the known values to find ∆H (Delta H):

∆H = (0.036) (4200) (42)

∴ ∆H = 6350.4 J/mol = 6.35 kJ/mol

Result: The enthalpy change during the reaction is + 6.35 kJ/mol. Thus, the reaction is endothermic in nature.

FAQ

What is represented by delta H?

Delta H (∆H) represents enthalpy change, i.e., the amount of heat or energy exchanged between the system and its surroundings. If ∆H > 0 → The reaction is endothermic in nature. If ∆H < 0 → The reaction is exothermic in nature.

How to find ∆H (Delta H) of a reaction?

The enthalpy change (∆H) formula is: ∆H(reaction) = Σ n ∆Hf°(products) – Σ m ∆Hf°(reactants) Where, ∆H(reaction) = Enthalpy change for the reaction Σ n ∆Hf° (products) = Sum of the standard enthalpies of the products multiplied by the number of moles of each (n) in the balanced chemical equation. Σ m ∆Hf° (reactants) = Sum of the standard enthalpies of the reactants multiplied by the number of moles of each (m) in the balanced chemical equation.

What are the units of ∆H (Delta H)?

∆H is measured in Joules/mole (J/mol) or kilojoules/mole (kJ/mol).

What is the difference between ∆H and ∆H°?

∆H stands for enthalpy change, while ∆H° denotes the standard enthalpy change, i.e., Delta H at standard conditions of temperature and pressure (273.15 K and 1 atm, respectively).

What does q = n ∆H represent?

In q = n. ∆H: q = heat exchanged during the reaction ∆H (Delta H)  = enthalpy change n = number of moles of the substance