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Q3, , 1., , Define standard enthalpy of formation. Calculate the enthalpy of formation of, , benzene from data, , 2(q) PCO ig) +3H,0,), A,H? = -3266.0KI, , 15, CoHew + 3° we, , Coy + Or) > COr A,H® = -393 IKI, Hag+ 5 Ox) > H2Oq A, H® = -286.0KJ, , Ans. Ans. The standard enthalpy change for the formation of one mole of a, compound from its elements in their most stable states of aggregation (also, known as reference states) is called Standard Molar Enthalpy of Formation., , 15, A.H® = 6A, Hon, + 3A, Hc.0) — Ay Hoop 7 Aron, , =-3266KJ = 6 x — 393.1 + 3x — 286 —A,H)?, 0, , (CoHe) ~, = -3218 kJ/mol, HOTS QUESTIONS, , Why standard entropy of an elementary substance is not zero whereas standard, enthalpy of formation is taken as zero?, , Ans. A substance has a perfectly ordered arrangement only at absolute zero., Hence , entropy is zero only at absolute zero. Enthalpy of formation is the heat, change involved in the formation of one mole of the substance from its, elements. An element formed from it means no heat change., , The equilibrium constant for a reaction is one or more if AG° for it is less than, zero. Explain, , Ans. —AG° = RT In K, thus if AG° is less than zero. i.e., it is negative, then In, K will be positive and hence K will be greater than one., , . Many thermodynamically feasible reactions do not occur under ordinary, , conditions. Why?, , Ans. Under ordinary conditions, the average energy of the reactants may be, less than threshold energy. They require some activation energy to initiate the, reaction.