Q64 / 160 · JIPMER 2005 · Chemistry

$\mathrm{C}_{\text {(dia) }}+\mathrm{O}_2 \longrightarrow \mathrm{CO}_2 ; \Delta H=-395.4 \mathrm{~kJ} / \mathrm{mole}$ $\mathrm{C}_{(\mathrm{gr})}+\mathrm{O}_2 \longrightarrow \mathrm{CO}_2 ; \Delta H=-393.5 \mathrm{~kJ} / \mathrm{mole}$ $\mathrm{C}_{(\mathrm{gr})} \longrightarrow \mathrm{C}_{\text {(dia) }} ; \Delta H=$ ?

  • A
    -3.8
  • B
    -1.9
  • C
    +3.8
  • Correct answer
    D
    +1.9

Solution

$\mathrm{C}($ dia $)+\mathrm{O}_2 \rightarrow \mathrm{CO}_2$ $(\Delta \mathrm{H}=-395.4 \mathrm{~kJ} / \mathrm{mole})$...(1) $\mathrm{C}(\mathrm{gr})+\mathrm{O}_2 \longrightarrow \mathrm{CO}_2$ $(\Delta \mathrm{H}=-393.5 \mathrm{~kJ} / \mathrm{mole})$...(2) subtracting eq. (1) from (2), we have $\begin{aligned} & \mathrm{C}(\mathrm{gr}) \longrightarrow \mathrm{C}_{\text {(dia) }} ; \\ & \begin{aligned}(\Delta \mathrm{H} & =+1.9 \mathrm{~kJ} / \mathrm{mole}) \\ & \quad[\because \Delta \mathrm{H}=-393.5-(-395.4) \\ & \left.=+1.9 \mathrm{~kJ} \mathrm{~mol}^{-1}\right]\end{aligned}\end{aligned}$

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

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