Question

calculate the enthalpy of combustion (kj/mol) of methyl linoleate (c₁₉h₃₄o₂), an important component of many biodiesel fuels made from corn oil and methanol, given the following standard enthalpies of formation:
co₂(g) = -393.5 kj/mol
h₂o(l) = -285.8 kj/mol
c₁₉h₃₄o₂(l) = -645.7 kj/mol
answer:
kj/mol

Explanation:

Step1: Write the combustion reaction

$$\ce{C_{19}H_{34}O_{2}(l) + \frac{53}{2}O_{2}(g) -> 19CO_{2}(g) + 17H_{2}O(l)}$$

Step2: Use the formula $\Delta H_{\text{rxn}}^{\circ}=\sum n\Delta H_{\text{f,products}}^{\circ}-\sum m\Delta H_{\text{f,reactants}}^{\circ}$

For the reaction, $\Delta H_{\text{f, }O_{2}(g)}^{\circ} = 0\ \text{kJ/mol}$ (standard state of oxygen).
$$\Delta H_{\text{rxn}}^{\circ}=[19\times(- 393.5)+17\times(-285.8)]-[-645.7+\frac{53}{2}\times0]$$

Step3: Calculate the value

First calculate the product - side:
$19\times(-393.5)=-7476.5$; $17\times(-285.8)=-4858.6$
Sum of product - side: $-7476.5-4858.6=-12335.1$
Then calculate the reactant - side: $-645.7$
$$\Delta H_{\text{rxn}}^{\circ}=-12335.1-(-645.7)$$
$$\Delta H_{\text{rxn}}^{\circ}=-12335.1 + 645.7=-11689.4\ \text{kJ/mol}$$

Answer:

$-11689.4\ \text{kJ/mol}$