Question

using the thermodynamic information in the aleks data tab, calculate the standard reaction free energy of the following chemical reaction: \\(\ce{ch_{4}(g) + 2o_{2}(g) -> co_{2}(g) + 2h_{2}o(g)}\\) round your answer to zero decimal places. \\(\square\\) kj

Explanation:

Step1: Recall the formula for standard reaction free energy

The formula for the standard reaction free energy ($\Delta G^{\circ}$) is $\Delta G^{\circ}=\sum n\Delta G_f^{\circ}(\text{products})-\sum m\Delta G_f^{\circ}(\text{reactants})$, where $n$ and $m$ are the stoichiometric coefficients of products and reactants respectively, and $\Delta G_f^{\circ}$ is the standard free energy of formation.

Step2: Find the standard free energy of formation values

From thermodynamic data (ALEKS Data tab, typical values):

• $\Delta G_f^{\circ}(\text{CH}_4(g))=-50.72\ \text{kJ/mol}$
• $\Delta G_f^{\circ}(\text{O}_2(g)) = 0\ \text{kJ/mol}$ (element in standard state)
• $\Delta G_f^{\circ}(\text{CO}_2(g))=-394.36\ \text{kJ/mol}$
• $\Delta G_f^{\circ}(\text{H}_2\text{O}(g))=-228.57\ \text{kJ/mol}$

Step3: Calculate the sum for products

For products: $\text{CO}_2(g)$ and $2\text{H}_2\text{O}(g)$
Sum of $n\Delta G_f^{\circ}(\text{products})=1\times\Delta G_f^{\circ}(\text{CO}_2(g)) + 2\times\Delta G_f^{\circ}(\text{H}_2\text{O}(g))$
$=1\times(-394.36)+2\times(-228.57)$
$=-394.36 - 457.14=-851.5\ \text{kJ/mol}$

Step4: Calculate the sum for reactants

For reactants: $\text{CH}_4(g)$ and $2\text{O}_2(g)$
Sum of $m\Delta G_f^{\circ}(\text{reactants})=1\times\Delta G_f^{\circ}(\text{CH}_4(g))+2\times\Delta G_f^{\circ}(\text{O}_2(g))$
$=1\times(-50.72)+2\times0=-50.72\ \text{kJ/mol}$

Step5: Calculate $\Delta G^{\circ}$

$\Delta G^{\circ}=\sum n\Delta G_f^{\circ}(\text{products})-\sum m\Delta G_f^{\circ}(\text{reactants})$
$=-851.5-(-50.72)$
$=-851.5 + 50.72=-800.78\ \text{kJ/mol}$ (rounded to zero decimal places: $-801\ \text{kJ}$)

Answer:

-801