Question

If the bond energies of H-H, Br- Br, and H-Br are 433,
192 and 364 kJ mol⁻¹ respectively, the ∆H° for the reaction
H₂ (g) + Br₂(g) → 2HBr(g) is
(a) – 261 kJ (b) + 103 kJ
(c) + 261kJ (d) – 103 kJ

Answer 1

The change in enthalpy is – 103 kJ.

Explanation:

• The change in enthalpy is defined as the amount of energy that is emitted off or absorbed into the system when substantial mole of the reactants react to form one mole of products.

• The bond energy of a compound is the amount of energy that is released when the bond is formed between two atoms.

• The enthalpy of the reaction is actually the energy that is absorbed or emitted from breaking the bonds.

• Here in the reaction, we can see that for every H-H and Br-Br bonds breaking, 2 H-Br bonds are formed.

• The enthalpy of the reaction = differences of the bond energy.

• Or, △H₀ =$433+192-364-364$

•             =-103KJ.

• So the enthalpy of the reaction is -103KJ.

For more information about enthalpy,

https://brainly.in/question/12853464

The enthalpy of a reaction at 273 K. is - 3.57 KJ.

what will be the enthalpy of reaction at 373 Kif AC

= zero :-

(1) - 3.57

(2) Zero

373

(3) – 3.57 x 373

(4) - 375​

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Answer 2

Answer:

-103kj/mol

-103kj/molUse this formula to solve thses type of questions

-103kj/molUse this formula to solve thses type of questions H°=Enthalpy of reactant-enthalpy of products.