Given: (A) 2CO(g) + O2(g) 2CO2(g), H1 = -x\,kJ mol-1 (B) C(graphite) + O2(g) CO2(g), H2 = -y\,kJ mol-1 The H for the reaction C(graphite) + 12O2(g) CO(g) is:

2y-x. 🧠 Target C + 1/2 O2 - CO = reaction (B) minus half of reaction (A) Hess's Law: combine the two given reactions to get the target. 🗺️ Derivation Take reaction (B): C + O2 - CO2, H = -y Take half of (A) reversed: CO2 - CO + 1/2 O2, H = +x/2 Adding the two steps: {C + O2 + CO2 - CO2 + CO + 1/2 O2} {C + 1/2 O2 - CO} H = -y + {x}{2} = {x - 2y}{2} With typical values (x 566 kJ for 2CO combustion, y 393.5 kJ for C combustion): H = (566-787)/2 = -110.5 kJ — matching the known value for CO formation. ✓ {{Answer: (c)}}

JEE Main · 2023 · Shift-ImediumTHERMO-092

Given: (A) 2CO(g) + O2(g) 2CO2(g), H1 = -x\,kJ mol-1 (B) C(graphite) + O2(g) CO2(g), H2 = -y\,kJ mol-1 The H for the…

Thermodynamics · Class 11 · JEE Main Previous Year Question

Question

Given: (A) $2\text{CO}(g) + \text{O}_2(g) \rightarrow 2\text{CO}_2(g)$ , $\Delta H_1^\circ = -x\,\text{kJ mol}^{-1}$ (B) $\text{C}(\text{graphite}) + \text{O}_2(g) \rightarrow \text{CO}_2(g)$ , $\Delta H_2^\circ = -y\,\text{kJ mol}^{-1}$

The $\Delta H^\circ$ for the reaction $\text{C}(\text{graphite}) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{CO}(g)$ is:

Options

a
$\frac{2x-y}{2}$
b
$\frac{x+2y}{2}$
c
$\frac{x-2y}{2}$
d
$2y-x$
✓

Correct Answerd

$2y-x$

Detailed Solution

🧠 Target $\ce{C + 1/2 O2 -> CO}$ = reaction (B) minus half of reaction (A) Hess's Law: combine the two given reactions to get the target.

🗺️ Derivation Take reaction (B): $\ce{C + O2 -> CO2}$ , $\Delta H = -y$

Take half of (A) reversed: $\ce{CO2 -> CO + 1/2 O2}$ , $\Delta H = +x/2$

Adding the two steps: $\ce{C + O2 + CO2 -> CO2 + CO + 1/2 O2}$ $\Rightarrow \ce{C + 1/2 O2 -> CO}$

$\Delta H = -y + \frac{x}{2} = \frac{x - 2y}{2}$

With typical values ( $x \approx 566$ kJ for $2\ce{CO}$ combustion, $y \approx 393.5$ kJ for $\ce{C}$ combustion): $\Delta H = (566-787)/2 = -110.5$ kJ — matching the known value for CO formation. ✓

$\boxed{\text{Answer: (c)}}$

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