Given (A) $2 \mathrm{CO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{CO}_{2}(\mathrm{~g}) \quad \Delta \mathrm{H}_{1}^{\theta}=-\mathrm{x} \mathrm{kJ} \mathrm{mol}^{-1}$ (B) C(graphite) $+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g}) \Delta \mathrm{H}_{2}^{\theta}=-\mathrm{ykJ} \mathrm{mol}^{-1}$ The $\Delta \mathrm{H}^{\theta}$ for the reaction $\mathrm{C}($ graphite $)+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}(\mathrm{g})$ is

JEE Main · 2023 · 10 Apr 2023 · Shift 1 · Chemical Thermodynamics

Given
(A)

2 CO (g) + O_{2} (g) \to 2 CO_{2} (g) Δ H_{1}^{θ} = - x kJ mol^{- 1}

(B) C(graphite)

+ O_{2} (g) \to CO_{2} (g) Δ H_{2}^{θ} = - ykJ mol^{- 1}

The

Δ H^{θ}

for the reaction

C (

graphite

) + \frac{1}{2} O_{2} (g) \to CO (g)

(A) $\frac{x - 2 y}{2}$ ✓ Correct
(B) $\frac{x + 2 y}{2}$
(C) $\frac{2 x - y}{2}$
(D) $2 y - x$

Answer: (A)

Solution

The target change, turning graphite and half a mole of oxygen into carbon monoxide, is built from the two data reactions by Hess's law. Burning graphite completely to carbon dioxide releases

- y

, and reversing half of the carbon monoxide combustion, that is

CO_{2} \to CO + \frac{1}{2} O_{2}

, costs

+ \frac{x}{2}

. Adding these two steps cancels the carbon dioxide and leaves graphite plus half a mole of oxygen forming carbon monoxide, with enthalpy change

- y + \frac{x}{2} = \frac{x - 2 y}{2}

JEE Main · 2023 · 10 Apr 2023 · Shift 1 · Single correct option

Practise more: all JEE Main PYQs on Chemical Thermodynamics →

JEE Main · 2023 · 10 Apr 2023 · Shift 1 · Chemical Thermodynamics

Solution

More JEE Main questions on Chemical Thermodynamics