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Explain Enthalpy of atomization (Delta_(a) H^( Theta )). |
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Answer» Solution :To break one mole of bonds completely to OBTAIN atoms in the gas phase is CALLED enthalpy of atomization." `H_(2(g)) to 2H_((g)), Delta_(a) H^( THETA ) = 435.0 "kj/mol"` During this process H atoms are formed by breaking H-H bonds in dihydrogen. The enthalpy change in this process is known as enthalpy of atomization `Delta_(a) H^( Theta) `. The enthalpy of atomization is also the bond dissociation enthalpy. e.g., `CH_(4(g)) to C_((g)) + 4H_((g)) ,Delta_(a) H^( Theta ) = 1665 "kj/mol"` `Na_((s)) to Na_((g)) , Delta_(a) H^( Theta ) = 108.4 "kj/mol"` In this case, the enthalpy of atomization is same as the enthalpy of sublimation. |
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