If the bond energies of H - H , Br - Br and H - Br are 433,192 and 364 kJ mol -1 respectively, then Δ H° for the reaction H 2(g)+ Br 2(g) longrightarrow 2 HBr (g) is

Question

If the bond energies of H - H , Br - Br and H - Br are 433,192 and 364 kJ mol -1 respectively, then Δ H° for the reaction H 2(g)+ Br 2(g) longrightarrow 2 HBr (g) is (A) -261 kJ (B) +103 kJ (C) +261 kJ (D) -103 kJ

Tardigrade Admin · Accepted Answer

For reaction H 2(g)+ Br 2(g) longrightarrow 2 HBr (g) Δ H°=? On the basis of bond energies of H 2, Br 2 and HBr , Δ H of above is calculated as follows Δ H=-[2 × bond energy of HBr - (bond energy of H 2+ bond energy of ..Cl 2)] Δ H =-[2 ×(364)-(433)+192 kJ =-[728-(625)] kJ =-103 kJ

Q. If the bond energies of H−H,Br−Br and H−Br are 433,192 and 364kJmol−1 respectively, then ΔH∘ for the reaction H2​(g)+Br2​(g)⟶2HBr(g) is

-261 kJ

+103 kJ

+261 kJ

-103 kJ

For reaction H2​(g)+Br2​(g)⟶2HBr(g)ΔH∘=? On the basis of bond energies of H2​,Br2​ and HBr,ΔH of above is calculated as follows ΔH=−[2× bond energy of HBr− (bond energy of H2​+ bond energy of Cl2​)] ΔH=−[2×(364)−(433)+192kJ =−[728−(625)]kJ=−103kJ

Q. If the bond energies of $H - H, B r - B r$ and $H - B r$ are $433, 192$ and $364 k J m o l^{- 1}$ respectively, then $Δ H^{\circ}$ for the reaction
$H_{2} (g) + B r_{2} (g) ⟶ 2 H B r (g)$ is