Heat of Hydration Formula

Hydration is the process through which water hardens concrete. Hydration is a chemical event in which the cement’s major constituents create chemical bonds with water molecules, resulting in hydrates or hydration products. Aggregates are inert particles that are bound together by cement.

The process of exchanging water in the body is known as hydration. Drinking water, chewing ice pieces, eating water-rich meals, drinking other fluids, or feeding them orally or directly can all help. When your body doesn’t have enough water to operate properly, you’re dehydrated.

What is Heat of Hydration? 

The reaction of cement with water is exothermic which liberates a considerable quantity of heat, This liberation of heat is called heat of hydration. The heat is delivered by hydration of one mole of particles at a consistent pressure. The more the particle is hydrated, the more heat is delivered.

Hydration Reaction

In a carbon double bond, a hydration process occurs when hydrogen and hydroxyl ions are bonded to a carbon. In most cases, one reactant (typically an alkene or alkyne) interacts with water to produce ethanol, isopropanol, or 2-butanol (alcohols). The general chemical reaction of hydration is given as,

RRC = CH₂ + H₂O/acid  →  RRC – CH₂ – OH

For example, in industrial usage ethanol is manufactured by the hydration of ethene.

CH₂ = CH₂ (g)   +   H₂O(g)     ⇄      CH₃CH₂OH (g)

Different compounds at different rates liberate different quantities of heat. If cement product is added with retarders it alters the fast setting properties of C₃A. The fineness of cement also has some influence on the rate of development of heat. Normal cement generally produces 89-90 calories/gm in 7 days and 90-100 calories/gm in 28 days. Thus, the reaction involves in the evolution of heat is faster in the early period and continues indefinitely at decreasing rate. 

Formula for Heat of Hydration

The Heat of Hydration formula is characterized by how much energy is delivered when one mole of particles goes through hydration. The heat of hydration recipe is given by,

Heat of hydration = ΔH _solution – ΔH _{lattice energy}

Heat of hydration in Cement 

When concrete is blended in with water, an exothermic reaction happens where much heat is delivered. At lower temperatures, it could consume a long time of the day for a concrete hydration reaction to create. Increasing temperatures lead to a fast production of heat.

Cement and other added substances are combined as one, the outcome is an exothermic reaction including cement and water (hydration). Subsequently, the heating effect is considered. The heat evolved by chemical reactions with water, such as that evolved during the setting and hardening of Portland cement or the difference between the heat of solution of dry cement and that of partially hydrated cement.

Formula to Determine Heat of Hydration in cement:

H = H₁ – H₂ – 0.4 (t_h – 25.0)

where,

• H = heat of hydration of ignited cement, kJ/kg,
• H₁ = heat of solution of dry cement,
• H₂ = heat of solution of a partially hydrated sample, and 
• t_h = final calorimeter temperature at end of determination on the partially hydrated sample, °C.

In the event that the heat of hydration isn’t as expected controlled, particularly assuming it is in an enormous chunk or design, the heat can’t escape without any problem. The heat of hydration can make extremely high inward temperatures in the design. This is especially obvious during development in a warm climate, assuming a lot of cement is poured rapidly, or then again in the event that there is an exceptionally high concrete to water proportion.

On the off chance that the heat of hydration isn’t overseen during development, it can cause extension while the concrete is solidifying and relieving. A lot of heat, particularly when joined with fake or climate-related cooling during the relieving system, can make critical breaks. The development and withdrawal of the substantial, alongside the subsequent breaks, can bring about difficult issues with the primary honesty of the substantial.

Sample Problems

Question 1: The sodium chloride lattice enthalpy is ΔH for NaCl is 800 kJ/mol. To make 1 M NaCl the solution heat is +7.0kJ/mol. Determine the heat of hydration of Na⁺ and Cl^–, where the heat of hydration of Cl^– is -500kJ/mol.

Solution:

Given that,

Lattice energy = 800 kJ/mol

Heat of solution = 7.0kJ/mol

Heat of hydration of Cl^– = -500kJ/mol

Substitute the values in the given formula:

We know, 

Heat of hydration = (ΔH_solution – ΔH_{lattice energy})

= 7 – 800

= -793

Therefore, Heat of hydration of Na⁺ + Cl^– = -793

Heat of hydration of Na⁺ = -793 – (-500)

Hence, Heat of hydration of Na⁺ = -293

Problem 2: The heat of a solution of dry cement is 90, The heat of a solution partially hydrated of a sample is 100. Assume that the final Calorimeter temperature of the sample obtained is 110 °C, Determine the heat of hydration of the cement.

Solution: 

We know heat of hydration of the cement is

H = H₁ – H₂ – 0.4(th – 25.0)          

Let H₁ = 90

and H₂ = 100

where final calorimeter temperature is 110°C

So, 

H = 90-100-0.4(110-25.0)

= -44 

Therefore, the heat of hydration of the cement is -44. 

Problem 3: The enthalpy of the solution of anhydrous CuSO₄ is −17.0 kcal and that of CuSO₄.5H₂O is 3.0 kcal. Calculate the enthalpy of hydration of CuSO₄.

Solution:

As given, 

Enthalpy of solution of anhydrous CuSO₄ = -17.0 kcal,

Enthalpy of solution of CuSO₄.5H₂O = 3.0 kcal,

Enthalpy of hydration of CuSO₄ = -17.0 – 3.0 

= – 20 kcal/mol

Therefore, the enthalpy of hydration of CuSO₄ is -20 kcal/mol.

Problem 4: The lattice enthalpy of solid NaCl is 660 kJmol⁻¹ and the enthalpy of the solution is 4 kJmol⁻¹. If the hydration enthalpy of Na⁺ and Cl⁻ ions are in the ratio of 2:3.5, what is the enthalpy of hydration of chloride ions?

Solution:

Given that,

ΔH_lattice = 660 kJmol⁻¹

ΔH_solution = 4kJmol^-1

ΔH_hydration = −ΔH_solution + (−ΔH_lattice) 

= −4−660

= −664 kJ mol⁻¹

The hydration enthalpy of sodium and chloride ions in the ratio of 2: 3.5. 

So, 

2x + 3.5x =−664 

or

x = −121

The enthalpy of hydration of chloride ion is,

-121 × 3.5 = -423.5 kJ mol^-1

Problem 5: Enthalpies of solution of BaCl₂.2H₂O and BaCl₂ are 8 and −24.2 kJ mol⁻¹ respectively. Calculate the heat of hydration of BaCl₂ to BaCl₂.2H₂O.

Solution:

Given that, 

BaCl₂(s) + aq. ⟶ BaCl₂ aq.                       ΔH= −24.2 kJ/mol                               …..(1)

BaCl₂.2H₂O(s) + aq. ⟶ BaCl₂ (aq.) + 2H₂O                     ΔH= 8kJ/mol

BaCl₂ + 2H₂O ⟶ BaCl₂ + 2H₂O + aq.                                 ΔH= -8kJ/mol                            …..(2)

Adding equation (1) & (2), we have 

BaCl₂(s) + aq. +BaCl_2(aq.) + 2H₂O _(l)   ⟶  BaCl_2aq. + BaCl₂.2H₂O(_s) + aq.   

ΔH= (−24.2)+(−8)

BaCl₂(s) + 2H₂O_(l)  ⟶  BaCl₂.2H₂O 

ΔH= −32.2kJ/mol

Hence, the heat of hydration of anhydrous BaCl₂ is 32.2 kJ/mol.