1. Tardigrade
  2. Question
  3. Chemistry
  4. 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.

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Q. The enthalpy for the following reactions $\left(\Delta H^{\circ}\right)$ at $25^{\circ} C$ are given below
(i) $\frac{1}{2} H _{2}(g)+\frac{1}{2} O _{2}(g) \longrightarrow OH (g) \Delta H^{\circ}=-10.06 \,kcal$
(ii) $H _{2}(g) \longrightarrow 2 H (g) \,\,\, \Delta H^{\circ}=104.18\, kcal$
(iii) $O _{2}( g ) \longrightarrow 2 O ( g ) \,\,\, \Delta H^{\circ}=118.32\, kcal$
Calculate the $O - H$ bond energy in the hydroxyl radical.

IIT JEEIIT JEE 1981Thermodynamics

Solution:

$\Delta H^{\circ}=\Sigma BE$ (reactants) $-\Sigma BE$ (products)
$\Rightarrow -10.06=\frac{1}{2}(104.18)+\frac{1}{2}(118.32)- BE ( O - H )$
$BE ( O - H )=121.31 \,kcal$