Given NH 3 g -3 Cl 2 g - NCl 3 g -3 HCl - - H 1N 2 g -3 H 2 g - 2 NH 3 g - - H 2H 2 g - Cl 2 g - 2 HCl g - - H 3The heat of formation of NCI 3 in terms of - H 1- - H 2 and - H 3 isA- - H 1- H 2-2-3-2- H 3B- - H 1- H 2-2-3-2- H 3- - H 1- H 2-2-3-2- H 3D- - H 1- H 2-2-3-2- H 3

The correct option is A Δ H_1 nbsp;+ Δ H_22 32Δ H_3 NH_3(g) + 3Cl_2(g)→ NCl_3(g) + 3HCl ; Δ H_1 N_2(g) + 3H_2(g)→ 2NH_3(g) ; Δ H_2 H_2(g) + Cl_2(g)→ 2HCl_( ...

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Byju's Answer

Standard XII

Physics

Calorimetry

Given NH 3 g ...

Question

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

$Δ H_{1} + \frac{Δ H_{2}}{2} - \frac{3}{2} Δ H_{3}$

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$Δ H_{1} - \frac{Δ H_{2}}{2} + \frac{3}{2} Δ H_{3}$

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$Δ H_{1} - \frac{Δ H_{2}}{2} - \frac{3}{2} Δ H_{3}$

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$Δ H_{1} + \frac{Δ H_{2}}{2} + \frac{3}{2} Δ H_{3}$

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Solution

The correct option is A
$Δ H_{1} + \frac{Δ H_{2}}{2} - \frac{3}{2} Δ H_{3}$

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

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

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

--------------------------------------------------------------------------------------

$\frac{1}{2} N_{2 (g)} + \frac{3}{2} C l_{2 (g)} \to N C l_{3 (g)}; Δ H_{4}$

$Δ H_{4} = Δ H_{1} + \frac{Δ H_{2}}{2} + \frac{3}{2} Δ H_{3}$

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Similar questions

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 :

Q. Given,

The heat formation of

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

in terms of

Δ H_{1}, Δ H_{2}

and

Δ H_{3}

is:

Q. 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 l_{3}

in terms of

Δ H_{1}, Δ H_{2} and Δ H_{3}

is:

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. $Δ H_{1} = 3272, Δ H_{2} = 5251, Δ_{e g} H = + 48$

Q. Consider the following two reactions:
(i) Propene +

H_{2} \to

Propane;

Δ H_{1}

(ii) Cyclopropane +

H_{2} \to

Propane;

Δ H_{2}

Then,

| Δ H_{2}

Δ H_{1} |

will be:

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Given NH3(g)+3Cl2(g)→NCl3(g)+3HCl ; ΔH1 N2(g)+3H2(g)→2NH3(g); ΔH2 H2(g) + Cl2(g)→ 2HClg; ΔH3 The heat of formation of NCI3 in terms of ΔH1,ΔH2andΔH3 is

The correct option is A ΔH1+ΔH22−32ΔH3 NH3(g)+3Cl2(g)→NCl3(g)+3HCl;ΔH1 N2(g)+3H2(g)→2NH3(g);ΔH2 H2(g)+Cl2(g)→2HCl(g);ΔH3 -------------------------------------------------------------------------------------- 12N2(g)+32Cl2(g)→NCl3(g);ΔH4 ΔH4=ΔH1+ΔH22+32ΔH3

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