∆H° of the following reaction H₂(g) + I₂(g) → 2HI(g) is -3kJ/mol.
<h3>What is Bond Enthalpy? </h3>
The minimum amount of energy which is required to braak down or form the bonds in chemical reaction is known as bond enthalpy.
It can be calculated as:
∆Hrxn = sum of ∆H bond broken - sum. of ∆H of bond formed.
In order to Calculate ∆Hrxn for the given equation we have:
Bond energies in kJ/mol
- H—H = 436
- H—I = 295
- I—I = 151
Now, the given reaction is
H₂(g) + I₂(g) → 2HI(g)
Here, 1 mol of H₂ and 1 mole of I₂ breaks to form 2 moles of HI.
Therefore,
We know that,
∆Hrxn = B. E(H—H) + B. E(I—I) - 2B. E(H—I)
= 436 + 151 - 2× 295
= 436+ 151 - 590
∆Hrxn = -3kJ/mol.
Thus, from the above conclusion we can say that ∆Hrxn of the reaction H₂(g) + I₂(g) → 2HI(g) is -3kJ/mol.
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