Calculate the enthalpy change for the reaction. H2(g) + Br2(g) → 2B

1.	Calculate the enthalpy change for the reaction. H2(g) + Br2(g) → 2BHr(g) Given the bond enthalpies H2, Br2 and HBr are 435 kJ mol-1 and 368 kJ mol-1 respectively.
Answer» For the reaction, H₂ (g) + Br₂ (g) → 2HBr(g) For reactions, Bond enthalpy of 1 mole of H-H bond = 435 kJ mol^-1 Bond enthalpy of 1 mole of Br-Br bond = 192 KJ mol^-1 For products, Bond enthalpy of 2 moles of H-Br bond = −2 × 368 = −736 kJ mol⁻¹ ∴ Enthalpy changes for the reaction ∆_rH^Θ = 435 + 192 − 736 = −109 kJ mol⁻¹

Calculate the enthalpy change for the reaction. H2(g) + Br2(g) → 2BHr(g) Given the bond enthalpies H2, Br2 and HBr are 435 kJ mol-1 and 368 kJ mol-1 respectively.

Answer»

For the reaction,

H₂ (g) + Br₂ (g) → 2HBr(g)

For reactions,

Bond enthalpy of 1 mole of H-H bond = 435 kJ mol^-1

Bond enthalpy of 1 mole of Br-Br bond = 192 KJ mol^-1

For products,

Bond enthalpy of 2 moles of H-Br bond = −2 × 368

= −736 kJ mol⁻¹

∴ Enthalpy changes for the reaction

∆_rH^Θ = 435 + 192 − 736

= −109 kJ mol⁻¹

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Calculate the enthalpy change for the reaction. H2(g) + Br2(g) → 2BHr(g) Given the bond enthalpies H2, Br2 and HBr are 435 kJ mol-1 and 368 kJ mol-1 respectively.

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