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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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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.

  1. 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.
  2. Temperature for spontaneity: Given: ΔH=300 kJ mol−1\Delta H = 300\ \text{kJ mol}^{-1}, ΔS=0.2 kJ K−1mol−1\Delta S = 0.2\ \text{kJ K}^{-1}\text{mol}^{-1} (both assumed constant with temperature). For spontaneity, we need ΔG<0\Delta G < 0: ΔG=ΔH−TΔS<0\Delta G = \Delta H - T\Delta S < 0 T>ΔHΔS=3000.2=1500 KT > \frac{\Delta H}{\Delta S} = \frac{300}{0.2} = 1500\ K …

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