Question

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

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 the ALEKS Data tab (standard thermodynamic data), we need to find the $\Delta G_f^\circ$ values for each compound:

• For $\ce{TiCl4(g)}$: Let's assume the value is $\Delta G_{f,\ce{TiCl4(g)}}^\circ$ (e.g., from standard tables, $\Delta G_{f,\ce{TiCl4(g)}}^\circ=-726.7\ \text{kJ/mol}$)
• For $\ce{H2O(g)}$: $\Delta G_{f,\ce{H2O(g)}}^\circ=-228.6\ \text{kJ/mol}$
• For $\ce{TiO2(s)}$: $\Delta G_{f,\ce{TiO2(s)}}^\circ=-888.8\ \text{kJ/mol}$
• For $\ce{HCl(g)}$: $\Delta G_{f,\ce{HCl(g)}}^\circ=-95.3\ \text{kJ/mol}$

Step3: Calculate the sum of $\Delta G_f^\circ$ for products

Products are $\ce{TiO2(s)}$ (coefficient 1) and $\ce{HCl(g)}$ (coefficient 4).
$\sum n\Delta G_f^\circ(\text{products}) = 1\times\Delta G_{f,\ce{TiO2(s)}}^\circ+ 4\times\Delta G_{f,\ce{HCl(g)}}^\circ$
Substitute the values:
$= 1\times(-888.8\ \text{kJ/mol})+4\times(-95.3\ \text{kJ/mol})$
$= -888.8 - 381.2$
$= -1270\ \text{kJ/mol}$

Step4: Calculate the sum of $\Delta G_f^\circ$ for reactants

Reactants are $\ce{TiCl4(g)}$ (coefficient 1) and $\ce{H2O(g)}$ (coefficient 2).
$\sum m\Delta G_f^\circ(\text{reactants})=1\times\Delta G_{f,\ce{TiCl4(g)}}^\circ + 2\times\Delta G_{f,\ce{H2O(g)}}^\circ$
Substitute the values:
$=1\times(-726.7\ \text{kJ/mol})+2\times(-228.6\ \text{kJ/mol})$
$= -726.7- 457.2$
$= -1183.9\ \text{kJ/mol}$

Step5: Calculate $\Delta G^\circ$ for the reaction

$\Delta G^\circ=\sum n\Delta G_f^\circ(\text{products})-\sum m\Delta G_f^\circ(\text{reactants})$
$= (-1270\ \text{kJ/mol})-(-1183.9\ \text{kJ/mol})$
$= -1270 + 1183.9$
$= -86.1\ \text{kJ/mol}$ (Wait, maybe my assumed values were wrong. Let's use correct standard values. Let's check again with accurate data:

Correct standard values (from standard thermodynamic tables):

• $\Delta G_{f,\ce{TiCl4(g)}}^\circ=-726.7\ \text{kJ/mol}$
• $\Delta G_{f,\ce{H2O(g)}}^\circ=-228.57\ \text{kJ/mol}$
• $\Delta G_{f,\ce{TiO2(s)}}^\circ=-888.8\ \text{kJ/mol}$
• $\Delta G_{f,\ce{HCl(g)}}^\circ=-95.299\ \text{kJ/mol}$

Recalculate products:
$\sum n\Delta G_f^\circ(\text{products})=1\times(-888.8)+4\times(-95.299)$
$= -888.8 - 381.196$
$= -1269.996\ \text{kJ/mol}$

Reactants:
$\sum m\Delta G_f^\circ(\text{reactants})=1\times(-726.7)+2\times(-228.57)$
$= -726.7- 457.14$
$= -1183.84\ \text{kJ/mol}$

Now, $\Delta G^\circ= -1269.996-(-1183.84)$
$= -1269.996 + 1183.84$
$= -86.156\ \text{kJ/mol}\approx -86\ \text{kJ/mol}$ (rounded to zero decimal places)

Answer:

-86