Question of 30
Q.(i) State Hess's law.
(ii) For the following reaction at 298 K: 2X + Y → Z, ΔH = 300 kJ mol⁻¹ and ΔS = 0.2 kJ K⁻¹ mol⁻¹. At what temperature will the reaction become spontaneous, considering ΔH and ΔS to be constant over the temperature range? [1 + 2]
West Bengal WbchseWest Bengal HS First Year (WBCHSE Class XI) Annual Examination 2018Subjective· 3mImportance★★★★★est
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Start your 14-day free trial to unlock the full solution →Hess's law says enthalpy change depends only on initial and final states; combine ΔH and ΔS via ΔG = ΔH − TΔS and solve for the temperature at which ΔG turns negative.
- Hess's Law of Constant Heat Summation: the total enthalpy change for a chemical reaction is the same whether the reaction occurs in a single step or via several intermediate steps, as long as the initial reactants and final products are the same — a direct consequence of enthalpy being a state function.
- Temperature for spontaneity: Given: , (both assumed constant with temperature). For spontaneity, we need : …
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