Derive the relationship between ∆H and ∆U.

1.	Derive the relationship between ∆H and ∆U.
Answer» ∴ Heat absorbed at constant volume q_v = ∆U …(i) Where ∆U Change in internal energy but most of chemical reactions are carried out not at constant volume, but in flask under constant atmospheric pressure. So, at constant pressure, e.g., (i) becomes ∆U = q_p − p∆V …(ii) Where q_p = heat absorbed by the system −p∆V = Expansion work done by the system Or U₂ − U₁ = q_p − p (V₂ − V₁ ) On rearranging, we get q_p = (U₂ + pV₂ ) − (U₁ − pV₁ ) …(iii) Now, we know that enthalpy H = U + pV …..(iv) So, equation (iii) becomes, q_p = H₂ − H₁ = ∆H For finite changes at constant Pressure, equation (iv) Can be written as: ∆H = ∆U + p∆V .....…(v) Since P is constant, so equation (v) becomes ∆H = ∆U + p∆V ..............(vi) The equation (vi) shows the relationship between ∆H and ∆U.

Answer»

∴ Heat absorbed at constant volume

q_v = ∆U …(i)

Where ∆U Change in internal energy but most of chemical reactions are carried out not at constant volume, but in flask under constant atmospheric pressure. So, at constant pressure, e.g., (i) becomes

∆U = q_p − p∆V …(ii)

Where q_p = heat absorbed by the system

−p∆V = Expansion work done by the system

Or U₂ − U₁ = q_p − p (V₂ − V₁ )

On rearranging, we get

q_p = (U₂ + pV₂ ) − (U₁ − pV₁ ) …(iii)

Now, we know that enthalpy

H = U + pV …..(iv)

So, equation (iii) becomes,

q_p = H₂ − H₁ = ∆H

For finite changes at constant Pressure, equation

(iv) Can be written as:

∆H = ∆U + p∆V .....…(v)

Since P is constant, so equation (v) becomes

∆H = ∆U + p∆V ..............(vi)

The equation (vi) shows the relationship between ∆H and ∆U.

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