Q58 / 200 · NEET 2003 · Chemistry

For which one of the following equations is $\Delta \mathrm{H}_{\text {react }}^{\circ}$ equal to $\Delta \mathrm{H}_f^{\circ}$ for the product?

  • A
    $\mathrm{N}_{2(g)}+\mathrm{O}_{3(g)} \rightarrow \mathrm{N}_2 \mathrm{O}_3(g)$
  • B
    $\mathrm{CH}_{4(g)}+2 \mathrm{Cl}_{2(\mathrm{~g})} \rightarrow \mathrm{CH}_2 \mathrm{Cl}_{2(l)}+$ $2 \mathrm{HCl}_{(g)}$
  • Correct answer
    C
    $\mathrm{Xe}_{(g)}+2 \mathrm{~F}_{2(g)} \rightarrow \mathrm{XeF}_{4(g)}$
  • D
    $2 \mathrm{CO}_{(g)}+\mathrm{O}_{2(g)} \rightarrow 2 \mathrm{CO}_{2(g)}$

Solution

Enthalpies of formation of elementary substances Xe and \(\mathrm{F}_2\) are taken as zero. For (c) \(\Delta H_{\text {reaction }}^{\circ}\) \(=\Delta H_f^{\circ}\left(\mathrm{XeF}_4\right)-\left[\Delta H_f^{\circ}(\mathrm{Xe})+2 \Delta H_f^{\circ}\left(\mathrm{F}_2\right)\right]\) Thus, \(\Delta \mathrm{H}^{\circ}\) reaction \(=\Delta \mathrm{H}^{\circ}{ }_{\mathrm{f}}\left(\mathrm{XeF}_4\right)\)

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

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