Question

What will be the enthalpy of formation of HBr if hypothetical bond energy of H-H, Br-Br and H-Br are 300 kJ/mole, 100kJ/mole and 250kJ/mole respectively? Reaction is as under H, +Br2 = 2HBr-------- ( A) -50 kJ/mole C) -100 kJ/mole B) +50 kJ/mole D) +1000 kJ/mole
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Answer 1

Answer:

• The ΔH for the reaction is - 100 KJ/mol

Explanation:

We are given with following reaction,

$\sf H_{2}\;+\; Br_{2}\longrightarrow 2\;HBr \; \; \; \;\;\Delta H=\;?$

The hypothetical bond energy is also given, which is

1. Bond energy of H-H is 300 KJ/mol

2. Bond energy of Br-Br is 100 KJ/mol &

3. Bond energy of H-Br is 250 KJ/mol

From the formula we know that,

ΔH = ∑ bond energy (reactants) - ∑ bond energy (Products)

Substituting,

⇒ ΔH = [(H-H) + (Br-Br)] - [2 × (H-Br)]

⇒ ΔH = [300 + 100] - [2 × 250]

⇒ ΔH = 400 - 500

⇒ ΔH = - 100

⇒ ΔH = -100 KJ/mol

∴ The ΔH for the reaction is - 100 KJ/mol.

Note:-

1) 2 is multiplied to the bond energy of products because 2 is the stochiometric coefficient of HBr.

2) Negative sign of enthalpy of fusion indicate that it is an exothermic reaction.

Answer 2

$\underline{\purple{\ddot{\MasterRohith}}}$

Given: -

• H2 + Br2 =2HBr(reaction)

And hypothetical bond energy is ,

• Bonding energy of H-H=300kJ/mol
• Bonding energy of Br-Br=100kJ/mol
• Bonding energy H-Br= 250kJ/mol

To prove :-

• Delta H value in the reaction.

Explanation :-

• We know that,
• DeltaH= bond energy (reactants)-bond energy of (Products)

• Now we substitute all the values in Given.

DeltaH= [(H-H)+(Br-Br)-(2×(H-Br)]

DeltaH= [(300+100)-(2×250)]

DeltaH= 400 - 500

DeltaH= -100kJ/mol.

Hence,

• DeltaH value for the reaction is-100kJ/mol^-1.

And this negative value indicates exothermic reaction that we should know first.

Hope it helps u mate .

Thank you .

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