# PPT 5.1 Measuring energy changes

Explore the PPT 5.1 Measuring energy changes in the above slider and also the notes below. Thermodynamics considers the entire universe in two parts:

System and Surrounding

1. System: Anything under observation during chemical reaction

Example: A cup of hot tea

• Surrounding: Everything apart from the system

Example: Entire space of the universe surrounding the cup of hot tea

Heat is exchanged between system and surrounding during a chemical reaction because bonds of reactants break and new bonds of products are formed.

According to the law of conservation of energy, the energy is conserved during any reaction.

Endothermic reactions: Heat is absorbed by the system. In endothermic reactions the energy required to break the bonds of reactants is greater than the energy released by the formation of new bonds in products.

Example: Thermal decomposition of calcium carbonate

Exothermic reactions: Heat is liberated by the system. In exothermic reactions the energy required to break the bonds of reactants is lesser than the energy released by the formation of new bonds in products.

Example: Burning of fuels

Heat is a form of energy measured in Joule or Kilo Joule and it is observed by change in temperature. Temperature is a measure of degree of hotness or coldness. It is measured in Kelvin or Celsius.

Kelvin Scale = Celsius + 273.15

Enthalpy(H): It is the stored internal energy in a chemical substance. It is not measurable however the change in enthalpy(∆H) is measurable at constant pressure.

∆H is positive when heat is absorbed by the system.

∆H is negative when heat is released from the system into the surroundings.

Standard conditions:

• Pressure 100kPa
• Concentration of 1 mol dm-3 for solutions
• All substances in their standard states
• Temp 298K

Standard Enthalpy of reaction(ΔHθ ) is the enthalpy change during a particular reaction occurred at standard conditions.

Example: Standard Enthalpy of Combustion(ΔHθC ), Standard Enthalpy of  formation(ΔHθf ),

Standard Enthalpy of Combustion(ΔHθC ): It is the enthalpy change  which takes place by the complete combustion of 1 mole of a substance under standard conditions in enough oxygen gas.

Example: CH4 (g)  +  2O2 (g)  —> CO2 (g)  +  2H2O (l)

ΔHθC = 890kJ/mol

Standard Enthalpy of  formation(ΔHθf ): It is the enthalpy change  which takes place when 1 mole of a substance is formed in its standard state from its constituent elements in their standard states.

2C (s) + 3H2 (g) + ½O2 (g) —>C2H5OH (l)

Specific Heat Capacity: It is the amount of heat energy required to heat 1 gram of a substance by 10C or 1K.

q= m C ΔT where q= heat produced, m= mass of water or sample solution, ΔT is the difference in temperature of the sample heated.

Example: Water has Specific Heat Capacity of 4.18 J g K which means  1 g water needs 4.18 J g K– energy to heat by 1K.

In all the calorimetry experiments, we assume that aqueous solutions have same density and specific Heat Capacity as that of water.

Calorimetry

It is the technique in which a substance is burned and the produced heat energy is used to heat the sample of water. We record the initial and final temperature and calculate the difference.

ΔT = T-T where T is the final temperature of the sample of water or solution while T is the initial temperature of the sample of water or solution.

The following equation is used to calculate the heat produced by the combustion of a substance.

• q= m C ΔT where q= heat produced, m= mass of water or sample solution, ΔT is the difference in temperature of the sample heated and C is the specific heat capacity.

You can calculate the enthalpy of combustion by using the following equation.

ΔH=-      where q= heat produced and n= moles of the substance combusted.

Example: In an experiment if 100 g water is heated by combustion of 5 g of ethanol to raise the temperature from 25 0C  to 45 0C. To calculate the enthalpy of combustion, we use the following steps.

ΔT=  T-T  = 45 – 25 = 20 0C

q= m C ΔT

q= 100 X 4.18 X 25

q= 10450 J

q= 10450/1000 = 10.45 KJ

n= 5g/46gmol = 0.109 mole

ΔH = 10.45/0.109

= 95.87 KJmol

I hope you have read PPT 5.1 Measuring energy changes.