The enthalpy of the reaction- H2g-12O2g- H2Og is - H1 and that of H2g-12O2g-H2Ol - H2 then-

The correct option is B Δ H_1 When H_2O (g) #160; changes to H_2O()l i.e. condensation takes place, heat is evolved. #160;i.e. ΔH_2 is great ...

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Enthalpy of Solution

The enthalpy ...

Question

The enthalpy of the reaction,
$H_{2 (g)} + \frac{1}{2} O_{2 (g)} \to H_{2} O_{(} g)$ is $Δ H_{1}$ and that of $H_{2 (g)} + \frac{1}{2} O_{2 (g)} \to {H_{2} O}_{(l)}$ $Δ H_{2}$ then:

Δ H_{1} < Δ H_{2}

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Δ H_{1} + Δ H_{2} = 0

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Δ H_{1} > Δ H_{2}

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Δ H_{1} = Δ H_{2}

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Solution

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

Δ H_{1}

Δ H_{2}

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

Given $N H_{3 (g)} + 3 C l_{2 (g)} \to N C l_{3 (g)} + 3 H C l$ ; $Δ H_{1}$

$N_{2 (g)} + 3 H_{2 (g)} \to 2 N H_{3 (g)}$ ; $Δ H_{2}$

$H_{2 (g)}$ + $C l_{2 (g)} \to$ $2 H C l_{g}$ ; $Δ H_{3}$

The heat of formation of $N C I_{3}$ in terms of $Δ H_{1}, Δ H_{2} a n d Δ H_{3}$ is

P r o p e n e + H_{2} ⟶ p r o p a n e; {Δ H}_{1}

C y c l o p r o p a n e + H_{2} ⟶ P o r p a n e; {Δ H}_{2}

{Δ H}_{2} - {Δ H}_{1}

H_{2} \to

Δ H_{1}

H_{2} \to

Δ H_{2}

| Δ H_{2}

Δ H_{1} |

List I	List II
A. Most reactive non-metal	1. $Δ H_{1} = 419, Δ H_{2} = 3051, Δ_{e g} H = - 48$
B. Most reactive metal	2. $Δ H_{1} = 1681, Δ H_{2} = 3374, Δ_{e g} H = - 328$
C. Least reactive elements	3. $Δ H_{1} = 738, Δ H_{2} = 1451, Δ_{e g} H = - 40$
D. Metal forming binary halide	4. $Δ H_{1} = 3272, Δ H_{2} = 5251, Δ_{e g} H = + 48$

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