At constant temperature and pressure, a correct statement for the reaction : $C O (g) + \frac{1}{2} O_{2} (g) \to C O_{2} (g)$ is:

Δ H = Δ E

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Δ H < Δ E

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Δ H > Δ E

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Δ H

is independent of the physical state of reactants

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Solution

The correct option is C $Δ H < Δ E$
$C O (g) + \frac{1}{2} O_{2} (g) \to C O_{2} (g)$

$Δ n_{g} = 1 - \frac{1}{2} - 1$ $= \frac{- 1}{2}$

$Δ H = Δ E + Δ n_{g} R T$

$Δ H = Δ E - \frac{1}{2} R T$

$∴ Δ E = Δ H + \frac{1}{2} R T$

$∴ Δ E > Δ H$

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Similar questions

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

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

Δ H

Δ E

C H_{4 (g)} + 2 O_{2 (g)} ⇌ C O_{2 (g)} + 2 H_{2} O_{(l)}

Which of the following statements is correct for the reaction $C O (g) + \frac{1}{2} O_{2} (g) \to C O_{2} (g)$ at constant temperature and pressure?

Δ H

k J

C (g) + O_{2} (g) \to C O_{2} (g)

H_{2} O (g) + C (g) \to C O (g) + H_{2} (g)

Δ H = + 131 k J

C O (g) + \frac{1}{2} O_{2} (g) \to {C O}_{2} (g)

Δ H = - 282 k J

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

Δ H = - 242 k J

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