Given, NH3(g) + 3Cl2(g) leftharpoons NCl3(g) + 3HCl(g) - Δ H1 N2(g) + 3H2(g) leftharpoons 2NH3(g) - Δ H2 H2(g) + Cl2(g) leftharpoons 2HCl(g) - Δ H3 The heat of formation of NCl3(g) in terms of Δ H1, Δ H2 and Δ H3 is

Question

Given, NH3(g) + 3Cl2(g) leftharpoons NCl3(g) + 3HCl(g) - Δ H1 N2(g) + 3H2(g) leftharpoons 2NH3(g) - Δ H2 H2(g) + Cl2(g) leftharpoons 2HCl(g) - Δ H3 The heat of formation of NCl3(g) in terms of Δ H1, Δ H2 and Δ H3 is (A)  Δ Hf = -Δ H1 + (1/2) Δ H2 - (3/2) Δ H3  (B)  Δ Hf = -Δ H1 + (1/2) Δ H2 + (3/2) Δ H3  (C)  Δ Hf = -Δ H1 - (1/2) Δ H2 - (3/2) Δ H3  (D) None of the above

Tardigrade Admin · Accepted Answer

NH3(g) + 3Cl2( g ) leftharpoons NCl3(g) + 3HCl(g); -Δ H1 ...(i) N2(g) + 3H2(g) leftharpoons 2NH3(g); -Δ H2...(ii) H2(g) + Cl2( g ) leftharpoons 2HCl( g ) ; + Δ H3...(iii) Multiply Eq. (i) by 2 and Eq. (iii) by 3, add Eq. (i) to Eq. (ii) and subtract Eq. (iii) to get the heat of formation of NCl3(g); N2 + 3Cl2 leftharpoons 2NCl3 2Δ Hf = 2 × ( -Δ H1) + ( -Δ H2) - 3 × ( +Δ H3) or Δ Hf = -Δ H1 -(1/2) Δ H2 - (3/2) Δ H3

Q. Given,
$N H_{3} (g) + 3 C l_{2} (g) ⇌ NC l_{3} (g) + 3 H Cl (g) - Δ H_{1}$
$N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g) - Δ H_{2}$
$H_{2} (g) + C l_{2} (g) ⇌ 2 H Cl (g) - Δ H_{3}$
The heat of formation of $NC l_{3} (g)$ in terms of $Δ H_{1}, Δ H_{2}$ and $Δ H_{3}$ is

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

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

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

None of the above

Q. Given, NH3​(g)+3Cl2​(g)⇌NCl3​(g)+3HCl(g)−ΔH1​ N2​(g)+3H2​(g)⇌2NH3​(g)−ΔH2​ H2​(g)+Cl2​(g)⇌2HCl(g)−ΔH3​ The heat of formation of NCl3​(g) in terms of ΔH1​,ΔH2​ and ΔH3​ is

ΔHf​=−ΔH1​+21​ΔH2​−23​ΔH3​

ΔHf​=−ΔH1​+21​ΔH2​+23​ΔH3​

ΔHf​=−ΔH1​−21​ΔH2​−23​ΔH3​

None of the above

Q. Given,
$N H_{3} (g) + 3 C l_{2} (g) ⇌ NC l_{3} (g) + 3 H Cl (g) - Δ H_{1}$
$N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g) - Δ H_{2}$
$H_{2} (g) + C l_{2} (g) ⇌ 2 H Cl (g) - Δ H_{3}$
The heat of formation of $NC l_{3} (g)$ in terms of $Δ H_{1}, Δ H_{2}$ and $Δ H_{3}$ is

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

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

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