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Prove Hess.s law of constant heat summation. |
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Answer» Solution :If a reaction takes place in several steps then its standard reaction enthalpy is the sum of the standard enthalpies of the intermediate reactions into which the overall reaction may be DIVIDED at the same temperature. e.g. : `C_((s)) + (1)/(2) O_(2(g)) to CO_((g)) , Delta H^( Theta) = (?)` Although `CO_((g))` is the major product, some `CO_2` gas is always produced in this reaction. Therefore, we cannot measure enthalpy change for the above reaction directly. Let us consider the following reactions : `C_((s)) + O_(2(g)) to CO_(2(g)) , Delta_(r) H^( Theta) = -393.5 kj//mol ""...(i)` `CO_((g)) + (1)/(2) O_(2(g)), Delta_(r) H^( Theta)= -283.0 kj//mol ""...(ii)` We can combine the above two reactions in such a way so as to obtain the desired reaction. We reverse equation (ii), `CO_(2(g)) to CO((g)) + (1)/(2) O_(2(g)), Delta_(r) H^( Theta) = + 283.0 kj//mol ""...(iii)` In GENERAL, if enthalpy of an overall reaction `Ato B` along one route is `Delta_(r) H and Delta_(r) H_(1), Delta_(r) H_(2), Delta_(r) H_(3) .....` representing enthalpies of reactions LEADING to same product, B along ANOTHER route, then we have, `Delta_(r) H= Delta_(r) H_(1) + Delta_(r) H_(2) + Delta_(r) H_(3)` |
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