Which of the following relationship is correct for the given exothermic reaction carried at constant pressure?

$C O (g) + 3 H_{2} (g) ⟶ C H_{4} (g) + H_{2} O (g)$

Δ E = Δ H

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Δ E > Δ H

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w < 0

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q > 0

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Solution

The correct option is B $Δ E > Δ H$

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Similar questions

C H_{4} (g) + H_{2} O (g) ⇌ C O (g) + 3 H_{2} (g), Δ H = + v e

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

(\frac{δ E}{δ T})_{P}

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Δ H - Δ E = 0 ?

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