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Q.a. i) State the first law of thermodynamics.

ii) Derive the relationship between ΔH and ΔU.
iii) Calculate the standard internal energy change for the reaction.
OF2(g) + H2O(g) -> O2(g) + 2HF(g) at 298K.
Given, standard enthalpies of formation in KJ mol-1 are
OF2(g) = +20, H2O(g) = -250 and HF(g) = -270. OR b. i) Define Gibbs energy and enthalpy.
ii) Calculate the enthalpy of formation of methane (CH4) from the following data:
a) C(s) + O2(g) -> CO2(g), ΔrH° = -393.5 KJmol-1
b) H2(g) + 1/2 O2(g) -> H2O(l); ΔrH° = -285.8 KJmol-1
c) CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l), ΔrH° = -890.3 KJmol-1
Nagaland NbseNagaland Board of School Education (Class XI) 2025Subjective· 5mImportance★★★★★
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First law: ΔU = q + w. ΔH = ΔU + ΔngRT. For the given reaction, ΔrH° = −310 kJ/mol and ΔU° ≈ −312.5 kJ/mol.

i) First law of thermodynamics: Energy can neither be created nor destroyed, only converted from one form to another; the total energy of an isolated system remains constant. For a system, the change in internal energy equals the heat supplied to the system plus the work done on the system: ΔU=q+w\Delta U = q + w.

ii) Relationship between ΔH and ΔU: Enthalpy is defined as H=U+PVH = U + PV. At constant pressure, for a finite change: ΔH=ΔU+PΔV\Delta H = \Delta U + P\Delta V. For a reaction involving ideal gases at constant T and P, PΔV=ΔngRTP\Delta V = \Delta n_g RT (where Δng\Delta n_g = moles of gaseous products − moles of gaseous reactants), giving:

ΔH=ΔU+ΔngRT\Delta H = \Delta U + \Delta n_g RT

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