Given-NH3g - 3Cl2g - NCl3g - 3HCl- - H1N2g - 3H2g - 2NH3g- - H2H2g - Cl2 g- 2HClg- - H3The heat of formation of NCl3 in terms of - H1- - H2 and - H3 is-

NH_3(g) + 3Cl_2 (g)→ NCl_3 (g)+ 3HCl ; Δ H_1...(i) N_2 (g)+ 3H_2(g) → 2NH_3(g) ; Δ H_2...(ii) H_2 (g)+ Cl_2(g) → 2HCl(g) ; Δ H_3...(iii) nbsp; 12N_2(g ...

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Byju's Answer

Standard XII

Chemistry

Limitations of First Law

Given,NH3g + ...

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 $N C l_{3}$ in terms of $Δ H_{1}, Δ H_{2} and Δ H_{3}$ 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 D
$Δ H_{1} + \frac{Δ H_{2}}{2} - \frac{3}{2} Δ H_{3}$

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

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

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

$\frac{1}{2} N_{2} (g) + \frac{3}{2} C l_{2} (g) \to N C l_{3} (g); Δ H_{4} . . . (i v)$
Equation $(i v)$ can be obtained by:
$(i) + (i i) \times \frac{1}{2} + (i i i) \times (- \frac{3}{2})$

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

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Similar questions

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:

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) ⇌ 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 that

N_{2} (g) + 3 H_{2} (g) \to 2 N H_{3} (g); △_{f} H^{⊖} = - 92 k J

, the standard molar enthalpy of formation in

k J m o l^{- 1}

N H_{3} (g)

is:

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

The correct option is D ΔH1+ΔH22−32ΔH3 NH3(g)+3Cl2(g)→NCl3(g)+3HCl; ΔH1...(i) N2(g)+3H2(g)→2NH3(g); ΔH2...(ii) H2(g)+Cl2(g)→2HCl(g); ΔH3...(iii) 12N2(g)+32Cl2(g)→NCl3(g); ΔH4...(iv) Equation (iv) can be obtained by: (i)+(ii)×12+(iii)×(−32) 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 $N C l_{3}$ in terms of $Δ H_{1}, Δ H_{2} and Δ H_{3}$ is: