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NH3g+3Cl2g⇌ N...

Question

$N H_{3} (g) + 3 C l_{2} (g) ⇌ N C l_{3} (g) + 3 H C l (g); - Δ H_{1}$
$N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g); Δ H_{2}$
$N_{2} (g) + C l_{2} (g) ⇌ 2 H C l (g); Δ H_{3}$
The enthalpy of formation of $N C l_{3}$ (g) in the terms of $Δ H_{1}$ , $Δ H_{2}$ and $Δ H_{3}$ is :

A

Δ H_{f} = Δ H_{1} + \frac{Δ H_{2}}{2} - \frac{3}{2} Δ H_{3}

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B

Δ H_{f} = - Δ H_{1} + \frac{Δ H_{2}}{2} - \frac{3}{2} Δ H_{3}

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C

Δ H_{f} = - Δ H_{1} - \frac{Δ H_{2}}{2} - \frac{3}{2} Δ H_{3}

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D

None

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Solution

The correct option is A $Δ H_{f} = - Δ H_{1} + \frac{Δ H_{2}}{2} - \frac{3}{2} Δ H_{3}$
Heat of formation of $N C l_{3}$ is the heat of the reaction of the reaction in which $N C l_{3}$ is formed from it's pure components- $N_{2}, C l_{2}$ .
Therefore the Reaction-
$(1 / 2) N_{2} + (3 / 2) C l_{2} \to N C l_{3}, δ H_{f} (N C l_{3})$ is obtained by-
Reaction(i) + Reaction(ii)/2 - (3/2)Reaction(iii)
So,
$Δ H_{f} = - Δ H_{1} + \frac{Δ H_{2}}{2} - \frac{3}{2} Δ H_{3}$

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Similar questions

The heat formation of

N {C l}_{3} (g)

Δ H_{1}, Δ H_{2}

Δ H_{3}

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

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 l_{3}

Δ H_{1}, Δ H_{2} and Δ H_{3}

Q. Consider the following reactions :

O_{(g)} + e^{-} \to O_{(g)}^{-}, Δ H_{1}

F_{(g)} + e^{-} \to F_{(g)}^{-}, Δ H_{2}

C l_{(g)} + e^{-} \to C l_{(g)}^{-}, Δ H_{3}

O_{(g)}^{-} + e^{-} \to O_{(g)}^{2 -}, Δ H_{4}

The correct statement(s) is/are

Match the correct ionization enthalpies and electron gain enthalpies of the following elements.

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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