Question

Calculate delta H for the reaction H(g) + Br(g) = HBr(g), given the following information:

H2(g) + Br2(g) = 2HBr(g) delta H = -72 kJ

H2(g) = 2H(g) delta H = +436 kJ

Br2(g)= 2Br(g) delta H = +224 kJ

Answer 1

alculate delta H for the reaction H(g) + Br(g) = HBr(g), given the following information:

H2(g) + Br2(g) = 2HBr(g) delta H = -72 kJ

H2(g) = 2H(g) delta H = +436 kJ

Br2(g)= 2Br(g) delta H = +224 kJ

We are given with,

i)             H₂ (g) ------> 2H (g)   ΔH = +436 kJ

Hence, 2H (g) ------> H₂ (g), (ΔH₁) = - ΔH = -436 kJ…………..(1)

ii)            Br₂ (g) ---- > 2Br (g)   ΔH = +224 kJ

Hence, 2Br (g) ---- > Br₂ (g) (ΔH₂) = - ΔH = -224 kJ…………….(2)

iii)       H₂ (g) + Br₂ (g) ------> 2HBr (g)   (ΔH₃) = -72 kJ………….(3)

By adding eq.(1), (2) and (3). Enthalpy is an Additive property.

2H (g) + 2Br (g) + H₂ (g) + Br₂ (g) ------> H₂ (g) + Br₂ (g) + 2HBr (g)    and      (ΔH₁) +(ΔH₂) + (ΔH₃)

Common things cancelled form both sides,

2H (g) + 2Br (g)   ------> 2HBr (g) and (ΔH₁) +(ΔH₂) + (ΔH₃)

Dividing throughout by 2 we get,

H (g) + Br (g)   ------> HBr (g) (ΔH_f) = [(ΔH₁) +(ΔH₂) + (ΔH₃)] /2

Hence,

(ΔH_f) = [(ΔH₁) +(ΔH₂) + (ΔH₃)] /2

(ΔH_f) = [(-436) +(-224) + (-72)] /2

(ΔH_f) = (-732)/2

(ΔH_f) = -366 kJ

Hence for, H (g) + Br (g)   ------> HBr (g), (ΔH_f) = - 366 kJ.

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