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The DeltaH_(f)^(0)N_(2)O_(5(g)) in KJ/mol on the basis of the following data is 2NO_((g))+O_(2(g))rarr2NO_(2(g))DeltaH_(f)^(0)=-114KJmol^(-1) NO_(2)+O_(2(g))rarr2N_(2)O_(5(g)) Delta_(f)H^(0)NO_((g))=90.2KJmol^(-1) |
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Answer» SOLUTION :`(1)/(2)N_(2(g))+(1)/(2)O_(2(g))rarrNO_((g))Delta_(F)H^(0)=90.2` `N_((g))+O_(2(g))rarr2NO_((g))Delta_(f)H^(0)=90.2xx2rarr(1)` `NO_((g))+O_(2(g))rarr2NO_(2(g))Delta_(f)H^(0)=-14rarr(2)` `2NO_(2(g))+(1)/(2)O_(2(g))rarrN_(2)O_(5(g))` `Delta_(f)H^(0)=(-102.6)/(2)=-51.3rarr(3)` From equations `(1)+(2)+(3)` `N_(2(g))+(5)/(2)O_(2(g))rarrN_(2)O_(5(g))` `Delta_(f)H^(0)N_(2)O_(5(g))=15.1KJmol^(-1)` |
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