Question

the standard enthalpy change of combustion to co2(g) and h2o(l) at 25°c of the organic solid dimethyl oxalate, c4h6o4(s), is determined to be -1658.5 kj/mol. what is the δhf° of c4h6o4(s) based on this value? use the following data: δhf°(h2o(l)) = -285.83 kj/mol, δhf°(co2(g)) = -393.51 kj/mol. δhf° = kj/mol

Explanation:

Step1: Write the combustion reaction

$C_4H_6O_4(s)+3.5O_2(g)
ightarrow4CO_2(g) + 3H_2O(l)$

Step2: Use the formula for $\Delta H^o_f$ of reaction

$\Delta H^o_{rxn}=\sum n\Delta H^o_f(products)-\sum m\Delta H^o_f(reactants)$. Here, for reactants $\Delta H^o_f(C_4H_6O_4(s))$ is unknown, $\Delta H^o_f(O_2(g)) = 0$ (standard - state element). For products, $n_{CO_2}=4$, $\Delta H^o_f(CO_2(g))=-393.51\ kJ/mol$, $n_{H_2O}=3$, $\Delta H^o_f(H_2O(l))=-285.83\ kJ/mol$ and $\Delta H^o_{rxn}=- 1658.5\ kJ/mol$.

Step3: Substitute values into the formula

$-1658.5\ kJ/mol=(4\times(-393.51\ kJ/mol)+3\times(-285.83\ kJ/mol))-\Delta H^o_f(C_4H_6O_4(s))$.
First, calculate the sum of $\Delta H^o_f$ of products:
$4\times(-393.51\ kJ/mol)+3\times(-285.83\ kJ/mol)=-1574.04\ kJ/mol - 857.49\ kJ/mol=-2431.53\ kJ/mol$.
Then, solve for $\Delta H^o_f(C_4H_6O_4(s))$:
$\Delta H^o_f(C_4H_6O_4(s))=-2431.53\ kJ/mol + 1658.5\ kJ/mol=-773.03\ kJ/mol$.

Answer:

$-773.03$ kJ/mol