QJEEM23Ch64e5: JEE Main 2023 Chemistry - Class 11 JEEM23Ch64e5 Solution: Given .tg {border-collapse:collapse;border-spacing:0;} .tg…

The standard enthalpy change for the reaction

$\mathrm{C(graphite)+\frac{1}{2}O_2(g)\to CO(g)}$

can be calculated using Hess's Law, which states that the total enthalpy change for a reaction is the same whether it occurs in one step or in many steps.

If we modify the given reactions to reflect the formation of CO, we can add the modified reactions together to get the reaction of interest.

Consider the reactions:

(A) $\mathrm{2CO(g)+O_2(g)\to 2CO_2(g)} \quad \Delta H_1^0=-x~kJ~mol^{-1}$

(B) $\mathrm{C(graphite)+O_2(g)\to CO_2(g)} \quad \Delta H_2^0=-y~kJ~mol^{-1}$

First, divide reaction (A) by 2 to get the reaction for 1 mol of CO:

(A/2) $\mathrm{CO(g)+\frac{1}{2}O_2(g)\to CO_2(g)} \quad \Delta H_1^0=-\frac{x}{2}~kJ~mol^{-1}$

Then, subtract the divided reaction (A) from reaction (B) to get the reaction for the formation of CO:

$\mathrm{C(graphite)+\frac{1}{2}O_2(g)\to CO(g)}$

$\Delta H^0=\Delta H_2^0-\Delta H_1^0=-y-(-\frac{x}{2})= \frac{x}{2} - y~kJ~mol^{-1}$

Therefore, the correct answer is Option A: $\frac{x-2y}{2}$