Question

What is the enthalpy for reaction 1 reversed? reaction 1 reversed: 2co2 + 3h2o→c2h5oh + 3o2 express your answer numerically in kilojoules per mole?

Answer 1

Answer: The enthalpy of the given reaction is 1234.8kJ/mol.

Explanation: Enthalpy change of the reaction is the amount of heat released or absorbed in a given chemical reaction.

Mathematically,

$\Delta H_{rxn}=\Delta H_f_{(products)}-\Delta H_f_{(reactants)}$

For the given reaction:

$2CO_2(g)+3H_2O(g)\rightarrow C_2H_5OH(l)+3O_2(g)$

$H_f_{(CO_2)}=-393.5kJ/mol$

$H_f_{(H_2O)}=-241.8kJ/mol$

$H_f_{(C_2H_5OH)}=-277.6kJ/mol$

$H_f_{(O_2)}=0kJ/mol$

$\Delta H_{rxn}=\Delta H_f_{(C_2H_5OH)}+3\Delta H_f_{(O_2)}-[2\Delta H_f_{(CO_2)}+3\Delta H_f_{(H_2O)}]$

Putting values in above equation, we get:

$\Delta H_{rxn}=[-277.6+3(0)]-[2(-393.5)+3(-241.8)]kJ/mol$

$\Delta H_{rxn}=1234.8kJ/mol$