MCQ
Data given for the following reaction is as follows:

$\mathrm{FeO}_{(0)}+\mathrm{C}_{\text {(gaplike) }} \longrightarrow \mathrm{Fe}_{(0)}+\mathrm{CO}_{(\mathrm{g})}$

Substance

$\Delta \mathrm{H}^{\circ}$

$\left(\mathrm{kJ} \mathrm{mol}^{-1}\right)$

$\Delta \mathrm{S}^{\circ}$

$\left(\mathrm{J} \mathrm{mol}^{-1} \mathrm{~K}^{-1}\right)$

$\mathrm{FeO}_{(s)}$ $-266.3$ $57.49$
$\mathrm{C}_{\text {(graphite) }}$ $0$ $5.74$
$\mathrm{Fe}_{(s)}$ $0$ $27.28$
$\mathrm{CO}_{(\mathrm{g})}$ $-110.5$ $197.6$

The minimum temperature in $\mathrm{K}$ at which the reaction becomes spontaneous is ....... .

(Integer answer)

  • $964$
  • B
    $864$
  • C
    $96.4$
  • D
    $9.64$

Answer

Correct option: A.
$964$
a
$\mathrm{T}_{\min }=\left(\frac{\Delta^{0} \mathrm{H}}{\Delta^{0} \mathrm{~S}}\right)$

$\Delta^{0} \mathrm{H}_{\mathrm{rxn}}=\left[\Delta_{\mathrm{f}}^{0} \mathrm{H}(\mathrm{Fe})+\Delta_{\mathrm{f}}^{0} \mathrm{H}(\mathrm{CO})\right]-$

$=\left[\Delta_{\mathrm{f}}^{0} \mathrm{H}(\mathrm{FeO})+\Delta_{\mathrm{f}}^{0} \mathrm{H}\left(\mathrm{C}_{(\text {graphite })}\right)\right]$

$=[0-110.5]-[-266.3+0]$

$=155.8 \, \mathrm{~kJ} / \mathrm{mol}$

$\Delta^{0} \mathrm{~S}_{\mathrm{rxn}}=\left[\Delta^{0} \mathrm{~S}(\mathrm{Fe})+\Delta^{0} \mathrm{~S}(\mathrm{CO})\right]-$

$\left[\Delta^{0} \mathrm{~S}(\mathrm{FeO})+\Delta^{0} \mathrm{~S}\left(\mathrm{C}_{(\mathrm{graphite})}\right)\right]$

$=[27.28+197.6]-[57.49+5.74]$

$=161.65 \,\mathrm{~J} / \mathrm{mol}-\mathrm{K}$

$\mathrm{T}_{\min }=\frac{155.8 \times 10^{3} \,\mathrm{~J} / \mathrm{mol}}{161.65\, \mathrm{~J} / \mathrm{mol}-\mathrm{K}}=963.8\, \mathrm{~K}$

$\simeq 964 \,\mathrm{k}$ (nearest integer)

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