InterviewSolution
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InterviewSolution
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The thermite reaction used ofr welding of metals involves the reaction `2Al(s) + Fe_(2)O_(3)(s) rarr Al_(2)O_(3)(s) + 2Fe(s)` What is `Delta H^(@) ` at `25^(@)C` for this reaction ? Given that the standard heats of formation of `Al_(2)O_(3)` and `Fe_(2)O_(3)` are - 1675.7 kJ and - 828.4 kJ `mol^(-1)` respectively. |
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Answer» We are given (i) `2Al(s) + (3)/(2) O_(2)(g) rarr Al_(2)O_(3)(s) , Delta _(r) H^(@) = - 1675.7 k J mol^(-1)` (ii) `2Fe(s) + (3)/(2) O_(2)(g) rarr Fe_(2)O_(3)(s) , Delta _(r) H^(@) = - 828 .4kJ mol^(-1)` We aim at `2Al(s) + Fe_(2)O_(3) (s), rarr Al_(2)O_(3)(s) + 2Fe(s), Delta_(r) H^(@) = ?` Equation (i) - Equation (ii) gives `2Al(s) + Fe_(2)O_(3)(s) rarr Al_(2)O_(3)(s)+2Fe(s), Delta_(r)H= - 1675.7 - ( 828.4)` `= - 847.3 kJ mol^(-1)` Alternative Method We aim at `2Al(s) + Fe_(2)O_(3)(s) rarr Al_(2)O_(3) +2Fe(s)lt Delta_(r) H^(@) = ?` `Delta _(r) H = ` Sum of `Delta_(r) H^(@)` of products `-` Sum of `Delta_(r) H^(@) =?` `= [ Delta_(f) H^(@) (Al_(2)O_(3))+ 2 xx Delta_(f ) H^(@) (Fe)] - [ 2 xx Delta _(f)H^(@) (Al)+Delta_(f) H^(@) ( Fe_(2)O_(3))]` `= [ 1675.7 +0] - [ 0+ ( -828.4)]` `= -847.3 kJ mol^(-1)` |
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