Question

Determine the heat of reaction (ΔHrxn) for the reduction of acetaldehyde (CH3CHO) to ethanol (C2H5OH) by using heat of formation data:

Answer 1

Answer: The enthalpy of the reaction is coming out to be -111.6 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 $\Delta H$

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

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

The chemical equation for the reduction of acetaldehyde to ethanol follows:

$CH_3CHO(g)+H_2(g)\rightarrow CH_3CH_2OH(l)$

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

$\Delta H_{rxn}=[(1\times \Delta H^o_f_{(CH_3CH_2OH(l))})]-[(1\times \Delta H^o_f_{(CH_3CHO(g))})+(1\times \Delta H^o_f_{(H_2(g))})]$

We are given:

$\Delta H^o_f_{(CH_3CH_2OH(l))}=-277.6kJ/mol\\\Delta H^o_f_{(CH_3CHO(g))}=-166kJ/mol\\\Delta H^o_f_{(H_2(g))}=0kJ/mol$

Putting values in above equation, we get:

$\Delta H_{rxn}=[(1\times (-277.6))]-[(1\times (-166))+(1\times (0))]\\\\\Delta H_{rxn}=-111.6kJ$

Hence, the enthalpy of the reaction is coming out to be -111.6 kJ.