At standard conditions, if the change in the enthalpy for the following reaction is -109 kJ mol-1 H2(g)+Br2(g) → 2HBr(g) Given that bond energy of H2 and Br2 is 435 KJ mol-1 and 192 KJ mol-1 , respectively, what is the bond energy (in KJ mol-1) of HBr ?

Question

At standard conditions, if the change in the enthalpy for the following reaction is -109 kJ mol-1 H2(g)+Br2(g) → 2HBr(g) Given that bond energy of H2 and Br2 is 435 KJ mol-1 and 192 KJ mol-1 , respectively, what is the bond energy (in KJ mol-1) of HBr ? (A) 368 (B) 736 (C) 518 (D) 259

Tardigrade Admin · Accepted Answer

Δ H=∑(B.E) textReactants-∑ (B.E) textproudcts -109=[B.E.(H-H)+B.E(Br-Br)]-[2× B.E(H-Br)] -109=435+192-2× B.E(H-Br) B.E(H-Br)=(435+192+109/2)=368 KJ/mol

$368$

$736$

$518$

$259$

$Δ H = \sum (B . E)_{Reactants} - \sum (B . E)_{proudcts}$
$- 109 = [B . E ._{(H - H)} + B . E_{(B r - B r)}] - [2 \times B . E_{(H - B r)}]$
$- 109 = 435 + 192 - 2 \times B . E_{(H - B r)}$
$B . E_{(H - B r)} = \frac{435 + 192 + 109}{2} = 368 K J / m o l$

Q. At standard conditions, if the change in the enthalpy for the following reaction is −109kJmol−1 H2(g)​+Br2(g)​→2HBr(g)​ Given that bond energy of H2​ and Br2​ is 435KJmol−1 and 192KJmol−1 , respectively, what is the bond energy (in KJmol−1) of HBr ?

368

736

518

259

ΔH=∑(B.E)Reactants​−∑(B.E)proudcts​ −109=[B.E.(H−H)​+B.E(Br−Br)​]−[2×B.E(H−Br)​] −109=435+192−2×B.E(H−Br)​ B.E(H−Br)​=2435+192+109​=368KJ/mol

$368$

$736$

$518$

$259$

$Δ H = \sum (B . E)_{Reactants} - \sum (B . E)_{proudcts}$
$- 109 = [B . E ._{(H - H)} + B . E_{(B r - B r)}] - [2 \times B . E_{(H - B r)}]$
$- 109 = 435 + 192 - 2 \times B . E_{(H - B r)}$
$B . E_{(H - B r)} = \frac{435 + 192 + 109}{2} = 368 K J / m o l$