Question

using the table of standard formation enthalpies that youll find under the aleks data tab, calculate the reaction enthalpy of this reaction under standard conditions:
2ch₃oh(l)+3o₂(g)→2co₂(g)+4h₂o(l)
round your answer to the nearest kj.

kj

Explanation:

Step1: Recall the formula for reaction enthalpy

The formula for reaction enthalpy ($\Delta H_{rxn}^{\circ}$) using standard formation enthalpies ($\Delta H_{f}^{\circ}$) is $\Delta H_{rxn}^{\circ}=\sum n\Delta H_{f}^{\circ}(\text{products})-\sum m\Delta H_{f}^{\circ}(\text{reactants})$, where $n$ and $m$ are the stoichiometric coefficients.

Step2: Find the standard formation enthalpies

From the data table:

• $\Delta H_{f}^{\circ}(\text{CH}_3\text{OH}(l))=-239\ \text{kJ/mol}$
• $\Delta H_{f}^{\circ}(\text{O}_2(g)) = 0\ \text{kJ/mol}$ (since it is an element in its standard state)
• $\Delta H_{f}^{\circ}(\text{CO}_2(g))=-394\ \text{kJ/mol}$
• $\Delta H_{f}^{\circ}(\text{H}_2\text{O}(l))=-286\ \text{kJ/mol}$

Step3: Calculate $\sum n\Delta H_{f}^{\circ}(\text{products})$

For the products:

$$
LATEXBLOCK0
$$

Step4: Calculate $\sum m\Delta H_{f}^{\circ}(\text{reactants})$

For the reactants:

$$
LATEXBLOCK1
$$

Step5: Calculate the reaction enthalpy

$$
LATEXBLOCK2
$$

Answer:

$-1454$