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

Question

Given, NH3(g) <=> 3Cl2(g) +NCl3(g) + 3HCl(g); -Δ H1 N2(g) +3H2(g) <=> 2NH3(g); -Δ H2 H2(g) +Cl2(g) <=> 2HCl(g) ; Δ H3 The heat of formation of NCl3(g) in the terms of Δ H1, Δ H2 and Δ H3 is (A) Δ Hf=-Δ H1+(Δ H2/2)-(3/2)Δ H3 (B) Δ Hf=Δ H1+(Δ H2/2)-(3/2)Δ H3 (C) Δ Hf=-Δ H1-(Δ H2/2)-(3/2)Δ H3 (D) none of these

Tardigrade Admin · Accepted Answer

Find (1/2) N2+(3/2)Cl2 → NCl3 ; Δ Hf Multiply Eq. (ii) by 1/2 and add to Eq. (i), Eq.(iii) by 3/2 and subtract from Eq.(i) ; we get Δ Hf =-Δ H1-[-(Δ H2/2)+(3/2)Δ H3] =-Δ H1+(Δ H2/2)-(3/2)Δ H3

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

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

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

none of these

Q. Given, NHX3​X(g)​​​3ClX2​X(g)​​+NClX3​(g)+3HCl(g)​; −ΔH1​ NX2​X(g)​​+3HX2​X(g)​​​2NHX3​X(g)​​; −ΔH2​ HX2​X(g)​​+ClX2​X(g)​​​2HClX(g)​​; ΔH3​ The heat of formation of NCl3(g)​ in the terms of ΔH1​, ΔH2​ and ΔH3​ is

ΔHf​=−ΔH1​+2ΔH2​​−23​ΔH3​

ΔHf​=ΔH1​+2ΔH2​​−23​ΔH3​

ΔHf​=−ΔH1​−2ΔH2​​−23​ΔH3​

none of these

Find 21​N2​+23​Cl2​→NCl3​;ΔHf​ Multiply Eq. (ii) by 1/2 and add to Eq. (i), Eq.(iii) by 3/2 and subtract from Eq.(i) ; we get ΔHf​=−ΔH1​−[−2ΔH2​​+23​ΔH3​] =−ΔH1​+2ΔH2​​−23​ΔH3​

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

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

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