Question

The enthalpy change for the reaction H2+C2H4=C2H6 is........ The bond energies are H-H=103 C-H=99,C-C =80,C=C=145

Answer 1

Answer:

-30

Explanation:

Answer 2

The enthalpy change for this reaction is -30 kJ

Explanation:

The balanced chemical reaction is:

$H_2+C_2H_4\rightarrow C_2H_6$

The expression for enthalpy change is,

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

$\Delta H=[(n_{C_2H_4}\times B.E_{C_2H_4})+(n_{H_2}\times B.E_{H_2}) ]-[(n_{C_2H_6}\times B.E_{C_2H_6})]$

$\Delta H=[(1\times B.E_{C=C})+(4\times B.E_{C-H})}+(1\times B.E_{H-H}) ]-[(1\times B.E_{C-C})+(6\times B.E_{C-H})]$

where,

n = number of moles

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

$\Delta H=[(1\times 145)+(4\times 99)+(1\times 103)]-[(1\times 80)+(6\times 99)]$

$\Delta H=-30kJ$

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