Sales Toll Free No: 1-855-666-7446

# Enthalpy Calculator

Top
Enthalpy Calculator calculates the heat energy transferred during a process in a system at constant pressure. Enthalpy is the internal energy plus the product of volume and constant pressure of the system. It's an online tool which helps solving enthalpy problems.

Formula to calculate enthalpy :
Enthalpy ($Q_{p}$) = $\Delta E$ + P$\Delta V$

$\Delta E$ = $E_{p}$ - $E_{r}$
$\Delta V$ = $V_{p}$ - $V_{r}$
where,
$\Delta E$ = change in internal energy
$E_{p}$ = internal energy of products
$E_{r}$ = internal energy of reactants
$\Delta V$ = change in volume
$V_{p}$ = volume of products
$V_{r}$ = volume of reactants.

## Steps

Step 1 : Check the given values of change in internal energy, pressure and change in volume.

Step 2 : Input the given values into the calculator according to the formula given above.

## Problems

Below are the problems based on Enthalpy Calculator.

### Solved Examples

Question 1: Calculate the enthalpy of a reaction where the initial internal energy of the products is 180 joules with volume of 15 cubic meters and internal energy of reactants is 110 joules with volume of 8 cubic meters at a pressure of 150 pascals ?
Solution:
Given:
Internal Energy of Products($E_{p}$) = 180 joules
Internal Energy of Reactants($E_{r}$) = 110 joules
Volume of Products($V_{p}$) = 15 cubic meters
Volume of Reactants($V_{r}$) = 8 cubic meters
Pressure(P) = 150 pascals
Enthalpy($Q_{p}$) = ?

Enthalpy($Q_{p}$) = $\Delta E$ + P$\Delta V$
= Ep - Er + P(Vp - Vr)
= 180 J - 110 J + 150 Pa(15$m^3$ - 8$m^3$)
= 1120 J.

Therefore, Enthalpy of this system is 1120 Joules.

Question 2: If 210 joules, 180 joules, 25 cubic meters and 12 cubic meters are the internal energy and volume of products and reactants, then find the enthalpy at a pressure of 190 pascals ?
Solution:
Given:
Internal Energy of Products($E_{p}$) = 210 joules
Internal Energy of Reactants($E_{r}$) = 180 joules
Volume of Products($V_{p}$) = 25 cubic meters
Volume of Reactants($V_{r}$) = 12 cubic meters
Pressure(P) = 190 pascals
Enthalpy($Q_{p}$) = ?

Enthalpy($QV_{p}$) = $\Delta E$ + P$\Delta V$
= Ep - Er + P(Vp - Vr)
= 210 J - 180 J + 190 Pa(25 $m^{3}$ - 12 $m^{3}$)
= 2500 J.

Therefore, Enthalpy of this system is 2500 Joules.