Q23 / 180 · COMEDK 2013 · Chemistry

The enthalpies of formation of $\mathrm{Al}_{2} \mathrm{O}_{3}$ and $\mathrm{Cr}_{2} \mathrm{O}_{3}$ are $-1596 \mathrm{~kJ}$ and $-1134 \mathrm{~kJ}$ respectively. $\Delta H$ for the reaction, $2 \mathrm{Al}+\mathrm{Cr}_{2} \mathrm{O}_{3} \longrightarrow 2 \mathrm{Cr}+\mathrm{Al}_{2} \mathrm{O}_{3}$ is

  • A
    $-2730 \mathrm{~kJ}$
  • Correct answer
    B
    $-462 \mathrm{~kJ}$
  • C
    $-1365 \mathrm{~kJ}$
  • D
    $+2730 \mathrm{~kJ}$

Solution

$\Delta H_{\text {Reaction }}=\left(\Delta H_{f}\right)_{\text {Product }}-\left\langle\Delta H_{f}\right)_{\text {Reactant }}$ $=-1596-(-1134)=-462 \mathrm{~kJ}$

Hence, the correct answer is option: 'B'.

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