Ionization enthalpy of sodium $$=$$ $$492$$ $$kJ/mole$$;Electron gain enthalpy of chlorine $$=$$ $$-349$$ $$kJ/mole$$;Sublimation enthalpy of sodium $$=$$ $$+108$$ $$kJ/mole$$ ;Bond dissociation enthalpy of chlorine $$=$$ $$+243$$ $$kJ/mole$$ ;Change in enthalpy for the process, $$Nas+12Cl2g$$→$$Na+g+Cl-g$$ will be

Question

Ionization enthalpy of sodium $$=$$ $$492$$ $$kJ/mole$$;Electron gain enthalpy of chlorine $$=$$ $$-349$$ $$kJ/mole$$;Sublimation enthalpy of sodium $$=$$ $$+108$$ $$kJ/mole$$ ;Bond dissociation enthalpy of chlorine $$=$$ $$+243$$ $$kJ/mole$$ ;Change in enthalpy for the process, $$Nas+12Cl2g$$→$$Na+g+Cl-g$$ will  be

Infinity Learn · Accepted Answer

Nas→Nag,          ∆$$H=+108$$ kJ..............$$.1$$ Nag→$$Na+g+e$$, ∆$$H=+492$$ kJ...............$$.2$$ $$12Cl2g$$→Clg,      ∆$$H=+121.5$$ $$kJ/mole$$...$$.3$$ $$Clg+e$$→$$Cl-g$$,    ∆$$H=-349$$ kJ...............$$.4$$ Adding $$1$$ ,$$2$$, $$3and$$ $$4$$ $$Nas+12Cl2g$$→$$Na+g+Cl-g$$, ∆$$H=+372.5$$ kJ ;

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Ionization enthalpy of sodium = 492 kJ/mole;Electron gain enthalpy of chlorine = -349 kJ/mole;Sublimation enthalpy of sodium = +108 kJ/mole ;Bond dissociation enthalpy of chlorine = +243 kJ/mole ;Change in enthalpy for the process, Nas+12Cl2g→Na+g+Cl-g will be

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