The heat of formation of PCl5(s) will be: Given: 2P(s)+3Cl2(g) xrightarrow2PCl3(l); Δ H=-151.8 k cal PCl3(l)+Cl2(g) xrightarrowPCl5(s) ; Δ H=- 32.8 k cal

Question

The heat of formation of PCl5(s) will be: Given: 2P(s)+3Cl2(g) xrightarrow2PCl3(l); Δ H=-151.8 k cal PCl3(l)+Cl2(g) xrightarrowPCl5(s) ; Δ H=- 32.8 k cal  (A)  -108.7 text k cal  (B)  +108.7 text k cal  (C)  -184.6 text k cal  (D)  +184.6 text k cal

Tardigrade Admin · Accepted Answer

P(s)+(5/2)Cl2(g) xrightarrowPCl5(s); Δ H=? Given: 2P(s)+2Cl2(g) xrightarrow 2PCl3(l)+151.8 kcal ...(i) PCl3(l)+Cl2(g) xrightarrow PCl5(s)+32.8 kcal ...(ii) Divide eq. (i) by 2 and add with equation (ii), we get P(s)+(5/2)Cl2(g) xrightarrowPCl5(s) +( (151.8/2)+32.8 ) k cal ∴ Δ Hf of PCl5(s)=- (75.9+32.8) =- 108.7 k cal

Q. The heat of formation of $PC l_{5} (s)$ will be: Given: $2 P (s) + 3 C l_{2} (g) 2 PC l_{3} (l);$ $Δ H = - 151.8 k c a l$ $PC l_{3} (l) + C l_{2} (g) PC l_{5} (s);$ $Δ H = - 32.8 k c a l$

$- 108.7 k c a l$

$+ 108.7 k c a l$

$- 184.6 k c a l$

$+ 184.6 k c a l$

Q. The heat of formation of PCl5​(s) will be: Given: 2P(s)+3Cl2​(g)​2PCl3​(l); ΔH=−151.8kcal PCl3​(l)+Cl2​(g)​PCl5​(s); ΔH=−32.8kcal

−108.7kcal

+108.7kcal

−184.6kcal

+184.6kcal

Q. The heat of formation of $PC l_{5} (s)$ will be: Given: $2 P (s) + 3 C l_{2} (g) 2 PC l_{3} (l);$ $Δ H = - 151.8 k c a l$ $PC l_{3} (l) + C l_{2} (g) PC l_{5} (s);$ $Δ H = - 32.8 k c a l$

$- 108.7 k c a l$

$+ 108.7 k c a l$

$- 184.6 k c a l$

$+ 184.6 k c a l$