The enthalpy for the following reactions (Δ H°) at 25° C are given below (i) (1/2) H 2(g)+(1/2) O 2(g) longrightarrow OH (g) Δ H°=-10.06 kcal (ii) H 2(g) longrightarrow 2 H (g) Δ H°=104.18 kcal (iii) O 2( g ) longrightarrow 2 O ( g ) Δ H°=118.32 kcal Calculate the O - H bond energy in the hydroxyl radical.

Question

The enthalpy for the following reactions (Δ H°) at 25° C are given below (i) (1/2) H 2(g)+(1/2) O 2(g) longrightarrow OH (g) Δ H°=-10.06 kcal (ii) H 2(g) longrightarrow 2 H (g) Δ H°=104.18 kcal (iii) O 2( g ) longrightarrow 2 O ( g ) Δ H°=118.32 kcal Calculate the O - H bond energy in the hydroxyl radical. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Δ H°= Sigma BE (reactants) - Sigma BE (products) ⇒ -10.06=(1/2)(104.18)+(1/2)(118.32)- BE ( O - H ) BE ( O - H )=121.31 kcal

Q. The enthalpy for the following reactions (ΔH∘) at 25∘C are given below (i) 21​H2​(g)+21​O2​(g)⟶OH(g)ΔH∘=−10.06kcal (ii) H2​(g)⟶2H(g)ΔH∘=104.18kcal (iii) O2​(g)⟶2O(g)ΔH∘=118.32kcal Calculate the O−H bond energy in the hydroxyl radical.

ΔH∘=ΣBE (reactants) −ΣBE (products) ⇒−10.06=21​(104.18)+21​(118.32)−BE(O−H) BE(O−H)=121.31kcal

$Δ H^{\circ} = Σ BE$ (reactants) $- Σ BE$ (products)
$\Rightarrow - 10.06 = \frac{1}{2} (104.18) + \frac{1}{2} (118.32) - BE (O - H)$
$BE (O - H) = 121.31 k c a l$