Prove that ΔH=ΔU+ΔnRT. what is the condition under which ΔU=ΔH?

Advertisement Remove all ads

#### Solution

∵ΔH=ΔU+PΔV

ΔH=ΔU+P(V_{2}-V_{1})

ΔH=ΔU+PV_{2}-PV_{1}

But PV_{1} = n_{1}RT

PV_{2} = n_{2}RT

∴ΔH= ΔU+ n_{1}RT - n_{2}RT

ΔH= ΔU+RT(n_{1}-n_{2})

∴ΔH= ΔU+ΔnRT |

Condition

i) When reaction is carried out in a closed vessel, so that volume remain constant ΔV=0

ii) When reaction involves only solids or liquids or solutions but no gaseous reactant or product.

iii) When reaction involves gaseous reactants and product but their number of mole are equal.

Concept: Chemical Thermodynamics and Energetic - First Law of Thermodynamics

Is there an error in this question or solution?

#### APPEARS IN

Advertisement Remove all ads