Answer :

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_{p}internal energy changes from U_{1} to U_{2} and volume increase from V_{1} to V_{2}

∆U = U_{2} - U_{1}---(4)

∆V = V_{2} - V_{1} --- (5)

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

∆U = q_{p} - p∆V

U_{2} - U_{1}= q_{p} - p (V_{2} - V_{1})

q_{p} = U_{2} - U_{1}+ p (V_{2} - V_{1})

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} = H_{2} – H_{1}

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_{A}RT

pV_{B} = n_{B}RT

Thus, pV_{B}- pV_{A} = n_{B}RT - n_{A}RT

p(V_{B} – V_{A}) = RT (n_{B}- n_{A})

p∆V = ∆n_{g}RT --- (7)

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

Therefore,

∆H = ∆U + ∆n_{g}RT

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