JEE Main 2023 — Thermodynamics (C) Question with Solution

(A) $2\mathrm{CO}\left(\mathrm{g}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)\to 2{\mathrm{CO}}_2\left(\mathrm{g}\right) {\mathrm{\Delta H}}_1^{\mathrm{o}}=-\mathrm{x} \mathrm{kJ} {\mathrm{mol}}^{-1}$

(B) ${\mathrm{C}}_{\mathrm{graphite}}+{\mathrm{O}}_2\left(\mathrm{g}\right)\to {\mathrm{CO}}_2\left(\mathrm{g}\right) {\mathrm{\Delta H}}_2^{\mathrm{o}}=-\mathrm{y} \mathrm{kJ} {\mathrm{mol}}^{-1}$

If the equation is multiplied the value $\mathrm{n}$, then the enthalpy change value also multiplied by factor $\mathrm{n}.$ If the enthalpy is positive in one direction, it is negative in other direction.

Multiply equation $\mathrm{B}$ by $2$ and subtract equation $\left(\mathrm{A}\right)$ from it

$2{\mathrm{C}}_{\mathrm{graphite}}+{\mathrm{O}}_2\left(\mathrm{g}\right)\to 2\mathrm{CO}\left(\mathrm{g}\right) {\mathrm{\Delta H}}^{\circ }=\mathrm{x}-2\mathrm{y}$

${\mathrm{C}}_{\mathrm{graphite}}+\frac{1}{2}{\mathrm{O}}_2\to \mathrm{CO}; {\mathrm{\Delta H}}^{\circ }=\frac{\mathrm{x}-2\mathrm{y}}{2}$