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Heat of Reaction Formula

The Heat of Reaction or Enthalpy of Reaction is the change in the enthalpy value of a chemical reaction at a constant pressure. Also, it is a thermodynamic unit of measurement to calculate the amount of energy per mole. This energy may be released or produced in a reaction. This article will explain the meaning of heat of reaction as well as it will deal with the Heat of reaction Formula with examples. Let us learn it!

What is the Heat of Reaction?

We may define Heat of a chemical reaction as the heat evolved in the surroundings or absorbed during the happening of a chemical reaction taking place at constant pressure and temperature. We measure this heat Joule unit. Mostly heat transfer is taking place between the reacting systems as one medium and surrounding as the other medium.

Heat of Reaction formula

Source: en.wikipedia.org

It is important to note that the amount of heat energy before and after the chemical change remains the same. Thus, the heat lost or gained in a reacting system is equal to the heat lost or gained in the surrounding.

An exothermic reaction liberates the heat, the temperature of the reaction mixture increases. An endothermic reaction absorbs heat, the temperature of the reaction mixture decreases.

The Formula for Heat of Reaction:

Therefore, the heat of reaction formula is:

\(Q = m \times c \times \Delta T\)

Where,

m Mass of the medium,
c The specific heat capacity of the medium of reaction
Q Heat of reaction
\(\Delta T\) the difference in temperature of the medium.

Also, we have the equation as:

\(Heat of reaction = \Delta H(products) — \Delta H(reactants)\)

Here \(\Delta H\) represents the change in heat value.

Solved Examples for Heat of Reaction Formula

Q.1: Determine the heat change using Heat of Reaction Formula which accompanies the combustion of a chemical when a certain mass of the substance is burnt in air to raise the temperature of 200 g of water initially at \(28 ^{\circ}C to 42 ^{\circ} C\). Specific heat capacity of water is given as \(4.2J g^{-1} K{-1}.\)

Solution: Given parameters in the problem are as below,

  • m = 200 g
  • \(c = 4.2 J g^{-1} K{-1}.\)
  • \(\Delta T = 42 – 28\)
  • \(\Delta T = 14^{\circ} C or 14 K\)

According to this problem, a certain mass of ethanol is burnt to raise the temperature of the water. It means heat absorbed by water is evolved from the combustion reaction of ethanol.

Since heat loss in the combustion reaction is equal to the heat gain by water. Therefore, the quantity of heat changed will be:

\(Q = m \times c \times \Delta T\)

\(Q = 200 \times 4.2 \times 14\)

Therefore, Q = 11760 J

Q.2: If Sodium chloride is dissolved in 100g of water at 20^{\circ} C, the after proper stirring temperature will be \(16^{\circ} C\). Compute the heat change during the process of dissolution, if the specific heat capacity of the solution is  \(4.18 J g^{-1} K{-1}\).

Solution: Given parameters are,

  • m= 100g
  • \(c= 4.18 J g^{-1} K{-1}.\)
  • \(\Delta T = 20 – 16\)
  • \(\Delta T = 4 K\)

Since heat absorbed by the salt will be the same as Heat lost by water. Therefore We have the formula,

\(Q = m \times c \times \Delta T\)

\(Q = 100 \times 4.18 \times 41\)

Therefore, Q = 1672 J

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