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Q.(i) State and explain Hess's law. [2]

(ii) For the reaction at 298 K, 2A + B → C, ΔH = 400 KJ mol⁻¹ and ΔS = 0.2 KJ K⁻¹ mol⁻¹. Considering ΔH and ΔS to be constant over the temperature range, at what temperature will the reaction become spontaneous? [2]
(iii) For an isolated system, ΔU = 0, what will be ΔS? [1] OR
(i) Explain the following terms:
(a) Closed system
(b) Intensive properties
(c) Enthalpy of atomization [3]
(ii) State and explain the first law of thermodynamics. [2]
Haryana BsehBoard of School Education Haryana (Senior Secondary Part-I / Class 11) 2026Subjective· 5mImportance★★★★★
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(i) Hess's Law: total ΔH is path-independent. (ii) The reaction turns spontaneous above 2000 K. (iii) For an isolated system, ΔS≥0\Delta S \geq 0.

(i) Hess's Law: The total enthalpy change for a reaction is the same regardless of the number of steps or the pathway taken to go from reactants to products, as long as the initial and final states are the same. This follows because enthalpy is a state function — so if a reaction can be expressed as the sum of two or more other reactions, its overall ΔH\Delta H equals the sum of the ΔH\Delta H values of those steps.

(ii) Temperature for spontaneity of 2A+B→C2A+B\rightarrow C: Using ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S, the reaction becomes spontaneous (ΔG<0\Delta G < 0) once TΔS>ΔHT\Delta S > \Delta H. The threshold (where ΔG=0\Delta G = 0) is:

T=ΔHΔS=400 kJ/mol0.2 kJ K−1mol−1=2000 KT = \frac{\Delta H}{\Delta S} = \frac{400\ kJ/mol}{0.2\ kJ\,K^{-1}mol^{-1}} = 2000\ K

Since both ΔH\Delta H and ΔS\Delta S are positive, the reaction becomes spontaneous above 2000 K (the entropy term TΔST\Delta S then outweighs the unfavourable positive ΔH\Delta H).

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