Assume each reaction is carried out in an open container. For which reaction will be $Δ H = Δ U$ ?

H_{2} (g) + {B r}_{2} (g) \to 2 H B r (g)

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

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P {C l}_{5} (g) \to P {C l}_{3} (g) + {C l}_{2} (g)

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

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Solution

The correct option is A $H_{2} (g) + {B r}_{2} (g) \to 2 H B r (g)$
Since, for any reaction the relation between $Δ H and Δ U$ is given by,
$Δ H = Δ U + (Δ n_{g}) R T$ where, $Δ n_{g} = change in gaseous moles$
Hence, the reaction in which the value of $Δ n_{g} = 0$ will be the one with $Δ H = Δ U$
Since in option $A$ , $Δ n_{g} = 2 - 1 - 1 = 0$
option $B$ , $Δ n_{g} = 2 + 1 - 2 = 1$
option $C$ , $Δ n_{g} = 1 + 1 - 1 = 1$
option $D$ , $Δ n_{g} = 2 - 2 - 1 = - 1$
Hence, answer is $A$ .

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