From the first law of thermodynamics

∆U = q + W --- (1)

∆U= internal energy

q= heat absorbed

W= work done

We know W = -p∆V --- (2)

Therefore we can write equation-1 as

∆U = q - p∆V--- (3)

Now if the system absorb q_pinternal energy changes from U₁ to U₂ and volume increase from V₁ to V₂

∆U = U₂ - U₁---(4)

∆V = V₂ - V₁ --- (5)

Substitute the value of ∆U and ∆V from equation-4 and equation-5 in equation-3 ie.

∆U = q_p - p∆V

U₂ - U₁= q_p - p (V₂ - V₁)

q_p = U₂ - U₁+ p (V₂ - V₁)

U, P, V are state functions and U + PV is heat content or enthalpy of the system which is denoted by

H = U + PV

So q_p = H₂ – H₁

q_p = ∆H

Therefore,∆H = ∆U + p∆V--- (6)

At constant temperature when the heat is absorbed, we measure the changes in enthalpy

∆H = q_p (when the heat is absorbed by the system at constant pressure)

∆H is positive for an endothermic reaction.

∆H is negative for an exothermic reaction.

At constant volume (∆V = 0)

∆U = q_v

Therefore,

∆H = ∆U = q_v

Now using ideal gas law if V_A is the total volume of gas reactant and V_B is the volume of gas product and n_A and n_B are moles of gas reactant and product respectively.

pV_A = n_ART

pV_B = n_BRT

Thus, pV_B- pV_A = n_BRT - n_ART

p(V_B – V_A) = RT (n_B- n_A)

p∆V = ∆n_gRT --- (7)

Put the value of p∆V from equation-7 to equation-5

Therefore,

∆H = ∆U + ∆n_gRT