Question

Calculate Δ H° for the reaction C 4H 4( g) + 2H 2( g) → C 4H 8( g), using the following data: Δ H° combustion for C 4H 4( g) = –2341 kJ/mol Δ H° combustion for H 2( g) = –286 kJ/mol Δ H° combustion for C 4H 8( g) = –2755 kJ/mol

Answer 1

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 $\Delta H^o$

The equation used to calculate enthalpy change is of a reaction is:  

$\Delta H^o_{rxn}=\sum [n\times \Delta H^o_{(product)}]-\sum [n\times \Delta H^o_{(reactant)}]$

For the given chemical reaction:

$C_4H_4(g)+2H_2(g)\rightarrow C_4H_8(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(1\times \Delta H^o_{(C_4H_8(g))})]-[(1\times \Delta H^o_{(C_4H_4(g))})+(2\times \Delta H^o_{(H_2(g))})]$

We are given:

$\Delta H^o_{(C_4H_8(g))}=-2755kJ/mol\\\Delta H^o_{(H_2(g))}=-286kJ/mol\\\Delta H^o_{(C_4H_4(g))}=-2341kJ/mol$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(1\times (-2755))]-[(1\times (-286))+(2\times (-2341))]\\\\\Delta H^o_{rxn}=2213kJ$

Hence, the enthalpy of the reaction is coming out to be 2231 kJ.