Question

A scientist measures the standard enthalpy change for the following reaction to be -574.9 kJ:

H2CO(g) + O2(g)CO2(g) + H2O(l)

Based on this value and the standard enthalpies of formation for the other substances, the standard enthalpy of formation of H2O(l) is

Answer 1

$H 2 ​ CO(g)+O 2 ​ (g)→CO 2 ​ (g)+H 2 ​ O(l)$

$ΔH rxn ​ =∑ΔH f 0 ​ (products)−∑ΔH f 0 ​ (reactants)$

$\Delta H^0_f(H_2CO(g)) = -116 \frac{kJ}{mol} ​$

$ΔH f 0 ​ (CO 2 ​ (g))=-393.5 kJ/mol$

$-574.9KJ=[1×ΔH f 0 ​ (H2O(l))+1×(-393.5 kJ/mol)]−[1×(−116)+1×0]$

$ΔH f 0 ​ (H2O(l)=297.4KJ$