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The enthalpy for the following reaction (∆Ho) at 25oC are given below :(i) 1/2 H2(g) + 1/2 O2(g) → OH(g) 10.06 kcal(ii) H2(g) → 2H(g)104.18 kcal(iii) O2(g) → 2O(g)118.32 kcalCalculate the O-H bond energy in the hydroxyl radical.

Shane Macguire , 10 Years ago
Grade upto college level
anser 1 Answers
Deepak Patra

Last Activity: 10 Years ago

The required reaction in terms of dissociation energy is OH(g) → O(g) + H(g); ∆H = ?
This equation can be achieved by (a) reversing the equation (i), (b) dividing equation (ii) and (iii) each by 2, and (c) adding the three resulting equations.
OH(g) → 1/2 H2(g) + 1/2 O2(g); ∆H = + 10.06 kcal [Reversing eq (i)]
1/2 H2(g) → H(g);p ∆H = - 52.09 kcal [ 1/2 Eq (ii)]
OH(g) → O(g) + H(g); ∆H = - 101.19 kcal (adding)
Thus one mole of OH(g) needs 101.19 kcal of energy to break into oxygen and hydrogen gaseous atom. Hence the bond energy of O-H bond is 101.19 kcal.

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