Given, H2(g) + Br2(g) → 2HBr(g), Δ H°1, and standard enthalpy of condensation of bromine is Δ H°2, standard enthalpy of formation of HBr at 25°C is

Question

Given, H2(g) + Br2(g) → 2HBr(g), Δ H°1, and standard enthalpy of condensation of bromine is Δ H°2, standard enthalpy of formation of HBr at 25°C is (A) Δ H°1/2 (B) Δ H°1/2+Δ H°2 (C) Δ H°1/2-Δ H°2 (D) (Δ H°1-Δ H°2)/2

Tardigrade Admin · Accepted Answer

H2(g)+Br2(g)→2HBr(g);Δ H=Δ H°1 ....(i) Br2(g)→ Br2(ℓ);Δ H=Δ H°2 ....(ii) Subtracting equation (ii) from equation (i), we get H2(g)+Br2(l)→2HBr(g);Δ H=Δ H°1-Δ H°2 Required equation, (1/2)H2(g)+(1/2)Br2(l)→ HBr(g); Δ H=[(Δ H°1=Δ H°2/2)]

Q. Given, $H_{2} (g) + B r_{2} (g) \to 2 H B r (g), Δ H_{1}^{\circ}$ , and standard enthalpy of condensation of bromine is $Δ H_{2}^{\circ}$ , standard enthalpy of formation of $H B r$ at $2 5^{\circ} C$ is

$Δ H_{1}^{\circ} /2$

$Δ H_{1}^{\circ} /2 + Δ H^{\circ} 2$

$Δ H_{1}^{\circ} /2 - Δ H^{\circ} 2$

$(Δ H_{1}^{\circ} - Δ H_{2}^{\circ}) /2$

Q. Given, H2​(g)+Br2​(g)→2HBr(g),ΔH1∘​, and standard enthalpy of condensation of bromine is ΔH2∘​, standard enthalpy of formation of HBr at 25∘C is

ΔH1∘​/2

ΔH1∘​/2+ΔH∘2

ΔH1∘​/2−ΔH∘2

(ΔH1∘​−ΔH2∘​)/2

H2​(g)+Br2​(g)→2HBr(g);ΔH=ΔH1∘​....(i) Br2​(g)→Br2​(ℓ);ΔH=ΔH2∘​....(ii) Subtracting equation (ii) from equation (i), we get H2​(g)+Br2​(l)→2HBr(g);ΔH=ΔH1∘​−ΔH2∘​ Required equation, 21​H2​(g)+21​Br2​(l)→HBr(g); ΔH=[2ΔH1∘​=ΔH2∘​​]

Q. Given, $H_{2} (g) + B r_{2} (g) \to 2 H B r (g), Δ H_{1}^{\circ}$ , and standard enthalpy of condensation of bromine is $Δ H_{2}^{\circ}$ , standard enthalpy of formation of $H B r$ at $2 5^{\circ} C$ is

$Δ H_{1}^{\circ} /2$

$Δ H_{1}^{\circ} /2 + Δ H^{\circ} 2$

$Δ H_{1}^{\circ} /2 - Δ H^{\circ} 2$

$(Δ H_{1}^{\circ} - Δ H_{2}^{\circ}) /2$