At constant temperature and pressure, which one of the following statement is correct for the reaction?
$C O (g) + \frac{1}{2} O_{2} (g) \to C O_{2} (g)$

Δ H > Δ U

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Δ H < Δ U

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Δ H = Δ U

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Δ H

is independent of the physical state of the reactants

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Solution

The correct option is B $Δ H < Δ U$
We know, $Δ H = Δ U + P Δ V = Δ U + Δ n_{g} R T$
where $Δ H$ = Change in enthalpy
$Δ U$ = Change in internal energy
$n_{g}$ = $n_{products} - n_{reactants}$
$n_{g}$ = $1 - 1.5 = - 0.5$
So, $Δ U > Δ H$

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Similar questions

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:

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)

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

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?

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