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Given the following equations calculate the standard enthalpy of the reaction: 2Fe_((s))+(3)/(2)O_(2_((g))) to Fe_(2)O_(3)(s)""DeltaH^(0)=? 2Al_((s))+Fe_(2)O_(3_((s))) to 2Fe_((s))+Al_(2)O_(3_((s)))""DeltaH^(0)=847.6kJ 2Al_((s))+(3)/(2)O_(2_((g))) to Al_(2)O_(3)(s)""DeltaH^(0)=-1670kJ |
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Answer» `2Fe_((s))+(3)/(2)O_(2_((g))) to Fe_(2)O_(3_(s))""DeltaH_(1)^(0)=?` Given equations: `(ii)2Al_((s))+Fe_(2)O_(3_((s))) to 2Fe_((s))+Al_(2)O_(3_((s)))""Delta_(2)^(0)=847.6kJ` (iii) `2Al_((s))+(3)/(2)O_(2_((g))) to Al_(2)O_(3_((s)))""DeltaH_(3)^(0)=-1670kJ` By substructing eq. (ii) From eq. (iii), we get eq. (i) `DeltaH_(1)^(0)=DeltaH_(3)^(0)-DeltaH_(2)^(0)` `=-1670-(847.6)` `=-822.4kJ` `therefore Delta_(R)H^(0)=DeltaH_(1)^(0)=-822.4kJ` |
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