JEE MAIN - Chemistry (2021 - 17th March Morning Shift - No. 21)
The standard enthalpies of formation of Al2O3 and CaO are $$-$$1675 kJ mol-1 and $$-$$635 kJ mol$$-$$1 respectively.
For the reaction
3CaO + 2Al $$ \to $$ 3Ca + Al2O3 the standard reaction enthalpy $$\Delta$$rH0 = _________ kJ.
(Round off to the Nearest Integer)
For the reaction
3CaO + 2Al $$ \to $$ 3Ca + Al2O3 the standard reaction enthalpy $$\Delta$$rH0 = _________ kJ.
(Round off to the Nearest Integer)
Answer
230
Explanation
$$3CaO + 2Al\mathrel{\mathop{\kern0pt\longrightarrow}
\limits_{}} 3Ca + A{l_2}{O_3}$$
$$\Delta H_{reaction}^o = 3\Delta {H^o}_f(Ca,s) + \Delta {H^o}_f(A{l_2}{O_3},s) - 3\Delta {H^o}_f(CaO,s) - 2\Delta {H^o}_f(Al,S)$$
$$ = 0 + ( - 1675) - 3( - 635) - 0$$
$$ = - 1675 + 1905$$
$$ = 230$$ KJ
$$\Delta H_{reaction}^o = 3\Delta {H^o}_f(Ca,s) + \Delta {H^o}_f(A{l_2}{O_3},s) - 3\Delta {H^o}_f(CaO,s) - 2\Delta {H^o}_f(Al,S)$$
$$ = 0 + ( - 1675) - 3( - 635) - 0$$
$$ = - 1675 + 1905$$
$$ = 230$$ KJ
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