Question

Calculate the enthalpy change for the reaction
H2(g)+ Br2(g)2HBr(g)Given :Bond enthalpy of H—H = 435 kJ mol-1Bond enthalpy of Br—Br = 192 kJ mol-1Bond enthalpy of H—Br = 364 kJ mol-1

Answer 1

Answer: -101 kJ

Explanation:-

The balanced chemical reaction is,

$H_2(g)+Br_2(g)\rightarrow 2HBr(g)$

The expression for enthalpy change is,

$\Delta H=\sum [n\times B.E(reactant)]-\sum [n\times B.E(product)]$

$\Delta H=[(n_{H_2}\times B.E_{H_2})+(n_{Br_2}\times B.E_{Br_2})]-[(n_{HBr}\times B.E_{HBr})]$

where,

n = number of moles

$B.E$ = Bond enthalpy

Now put all the given values in this expression, we get

$\Delta H=[(1\times 435+1\times 192)-(2\times 364)]$

$\Delta H=-101kJ$

Therefore, the enthalpy change for this reaction is -101 kJ.

Answer 2

Answer:

-102

Explanation: