Question

The enthalpies (AH) of formation of Al2O3 and Cr2O3 are - 1596 kj and
- 1134 kJ respectively. AH for the reaction : 2 Al + Cr2O3 = 2Cr + Al2O3.is :


(1) -462 kJ
(2) -1365 kJ
(3) -2730 kJ
(4) +2730 kJ
​

Answer 1

The standard enthalpy change for the given reaction is -462 kJ

Explanation:

The chemical equation is:

$2Al+Cr_2O_3\rightarrow 2Cr+Al_2O_3$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(2\times \Delta H^o_f_{(Cr})+(1\times \Delta H^o_f_{(Al_2O_3)})]-[(1\times \Delta H^o_f_{(Cr_2O_3)})+(2\times \Delta H^o_f_{(Al)})]$

We are given:

$\Delta H^o_f_{(Al_2O_3)}=-1596kJ/mol\\\Delta H^o_f_{(Cr_2O_3)}=-1134kJ/mol\\\Delta H^o_f_{(Cr)}=0kJ/mol\\\Delta H^o_f_{(Al)}=0kJ$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(2\times (0))+(1\times (-1596))]-[(1\times (-1134)})+(2\times (0))]$

$\Delta H^o_{rxn}=-462kJ$

Hence, the standard enthalpy change for the given reaction is -462 kJ

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