Respuesta :
Answer: The enthalpy of the reaction is coming out to be 2231 kJ.
Explanation:
Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as [tex]\Delta H^o[/tex]
The equation used to calculate enthalpy change is of a reaction is:
[tex]\Delta H^o_{rxn}=\sum [n\times \Delta H^o_{(product)}]-\sum [n\times \Delta H^o_{(reactant)}][/tex]
For the given chemical reaction:
[tex]C_4H_4(g)+2H_2(g)\rightarrow C_4H_8(g)[/tex]
The equation for the enthalpy change of the above reaction is:
[tex]\Delta H^o_{rxn}=[(1\times \Delta H^o_{(C_4H_8(g))})]-[(1\times \Delta H^o_{(C_4H_4(g))})+(2\times \Delta H^o_{(H_2(g))})][/tex]
We are given:
[tex]\Delta H^o_{(C_4H_8(g))}=-2755kJ/mol\\\Delta H^o_{(H_2(g))}=-286kJ/mol\\\Delta H^o_{(C_4H_4(g))}=-2341kJ/mol[/tex]
Putting values in above equation, we get:
[tex]\Delta H^o_{rxn}=[(1\times (-2755))]-[(1\times (-286))+(2\times (-2341))]\\\\\Delta H^o_{rxn}=2213kJ[/tex]
Hence, the enthalpy of the reaction is coming out to be 2231 kJ.
