If the bond energies of H - H , Br - B r and H - Br are 433,192 and 364 kJ mol -1 respectively, then Δ H° for the reaction, H2(g) +Br2 (g) → 2HBr (g) is

Question

If the bond energies of H - H , Br - B r and H - Br are 433,192 and 364 kJ mol -1 respectively, then Δ H° for the reaction, H2(g) +Br2 (g) → 2HBr (g) is (A) -261 kJ (B) +103 kJ (C) +621 kJ (D) -103 kJ

Tardigrade Admin · Accepted Answer

Required reaction H 2(g)+ Br 2(g) longrightarrow 2 HBr (g) Given, BE ( H - H )=433 kJ mol -1 BE ( Br - Br )=192 kJ mol -1 BE ( H - Br )=364 kJ mol -1 ∵ Δ HR° = Sigma BE (Reactants) - Sigma BE (Products) Δ HR° =(433+ text 192) -(2 × 364) Δ HR° =625-728 Δ HR° =-103 kJ mol -1

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,
$H_{2} (g) + B r_{2} (g) \to 2 H B r (g)$ is

$- 261 k J$

$+ 103 k J$

$+ 621 k J$

$- 103 k J$

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

−261kJ

+103kJ

+621kJ

−103kJ

Required reaction H2​(g)+Br2​(g)⟶2HBr(g) Given, BE(H−H)=433kJmol−1 BE(Br−Br)=192kJmol−1 BE(H−Br)=364kJmol−1 ∵ΔHR∘​=ΣBE (Reactants) -\Sigma BE (Products) ΔHR∘​=(433+ 192) −(2×364) ΔHR∘​=625−728 ΔHR∘​=−103kJmol−1

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,
$H_{2} (g) + B r_{2} (g) \to 2 H B r (g)$ is

$- 261 k J$

$+ 103 k J$

$+ 621 k J$

$- 103 k J$