Given that, $\begin{aligned} & \mathrm{C}(\mathrm{s})+\mathrm{O}_2(\mathrm{~g}) \longrightarrow…

Given that, $\begin{aligned} & \mathrm{C}(\mathrm{s})+\mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{CO}_2(\mathrm{~g}) ; \Delta H^{\circ}=-x \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & 2 \mathrm{CO}(\mathrm{g})+\mathrm{O}_2(\mathrm{~g}) \longrightarrow 2 \mathrm{CO}_2(\mathrm{~g}) ; \Delta H^{\circ}=-y \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned}$ The enthalpy of formation of $\mathrm{CO}$ will be
  1. $\frac{y-2 x}{3}$
  2. $\frac{y-2 x}{2}$
  3. $\frac{2 x-y}{2}$
  4. $\frac{x-y}{2}$

Solution

$\mathrm{C}(\mathrm{s})+\mathrm{O}_2(\mathrm{~g}) ightarrow \mathrm{CO}_2(\mathrm{~g})$ $\begin{aligned} & \Delta \mathrm{H}^{\circ}=-x \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & 2 \mathrm{CO}(\mathrm{g})+\mathrm{O}_2(\mathrm{~g}) ightarrow 2 \mathrm{CO}_2(\mathrm{~g}), \\ & \Delta \mathrm{H}^{\circ}=-y \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned}$ by applying eq. (i) $-\frac{1}{2} \times$ Eq. (ii) $\begin{aligned} & \mathrm{C}(\mathrm{s})+\frac{1}{2} \mathrm{O}_2 \mathrm{~g} ightarrow \mathrm{CO}(\mathrm{g}) \\ & \therefore \quad \Delta f \mathrm{H}(\mathrm{CO})=-x-\left(-\frac{y}{2}ight) \\ & =-x+\frac{y}{2}=\frac{y-2 x}{2} \end{aligned}$ *

Asked in: JEE-TOPICTESTS-CHEMISTRY

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