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Given the following equations calculate ...

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`

Text Solution

Verified by Experts

The correct Answer is:
Standard enthalpy of the reaction `=Delta_(r)H^(0)=-822.4kJ`
Required equation:
`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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