### Enthalpy Calculator:

Enter the aₙ coefficient in 2, Reactant A in SO₃(g), bₙ coefficient in 0, Reactant B in none, dₙ coefficient in 2, Product D in SO₂(g), eₙ coefficient in 1, Product E in O₂(g), and you get the results in Change in enthalpy in 197.78kJ.

Enthalpy formula is ΔH_{(j)} in joules is equal to the Q_{1}, Q_{2} are the internal energies of the products and reactants (Q_{2(j)} – Q_{1(j)} ) in joules and the product of the pressure p_{(pa)} in pascal and multiply the V_{1}, V_{2} are the volume of the products and reactants (V_{2(m3)} –V_{1(m3)} ) in meter^{3}. Hence the enthalpy formula can be written as

ΔH_{(j)}= (Q_{2(j)} – Q_{1(j)} ) + p_{(pa)} * (V_{2(m3)} –V_{1(m3)} )

or Simply You can write it as

ΔH = Q + P * V

### Derivation:

ΔH = Q + PV

Here,

Q = Internal energy in Joules

P = Pressure in Pascal

V = Volume in Cubic Meter

ΔH = Enthalpy in Joules

During isobaric condition, the pressure will be constant, then enthalpy of the system will be

ΔH= (Q_{2} – Q_{1} ) + P * (V_{2} –V_{1} )

Or simply

ΔH= Δ Q + P * Δ V

Where,

Q_{1}, Q_{2} in joules and they are internal energies of the substances

V_{1}, V_{2} Volume

ΔH = change in enthalpy in joules

ΔV = change in volume in cubic meter

### Standard Enthalpy:

Δ H ^{0 }_{reaction =} Σ_{ }Δ H ^{0 }_{F(products) } – Σ Δ H ^{0 }_{f(reactant)}

Where,

Σ Δ H ^{0 }_{f(reactant) }sum of the standard enthalpies – reactants in joules

Σ Δ H ^{0 }_{f(products)} sum of the standard enthalpies – products in joules

Δ H ^{0 }_{reaction} enthalpy change in joules

### Example:1

Calculate the enthalpy of products which one has the internal energy of the 98 joules and for the second product has the internal energy of 98 joules and internal volume V1 = 55 m^{3 } and V2 = 100 m^{3} and pressure P = 65 pa.

### Answer:

Q_{1 }= 98 joule

V_{1}= 55 meter.cubic

Q_{2} = 98 joules

V_{2} =100 meter.cubic

P = 65 pascal

Using formula

ΔH_{(j)}= (Q_{2(j)} – Q_{1(j)} ) + p_{(pa)} * (V_{2(m3)} –V_{1(m3)} )

=(98-98)+65 *(55-100) = 2925 joules

Therefore change in enthalpy ΔH_{(j)}= 2935 joules

### Example:2

Calculate the change in enthalpy of the water formation whose reaction 2H_{2} + O _{2} → 2 H_{2}O ?

### Answer:

2H_{2} + O _{2} → 2 H_{2} O

The standard enthalpy of formation of substances is here

H_{2} =0,

O_{2} =0,

H_{2 } o = -285.8

Change in enthalpy formula is

Δ H ^{0 }_{reaction =} Σ_{ }Δ H ^{0 }_{F(products) } – Σ Δ H ^{0 }_{f(reactant)}

Σ Δ H ^{0 }_{f(reactant)} =2(0) +0 =0

Σ_{ }Δ H ^{0 }_{F(products) } =2(-285.8)=571.7

Δ H ^{0 }_{reaction} =-517.7 -0 = -517.7

Therefore, change in enthalpy of the reaction is -517 .7 kj