Question

using the thermodynamic information in the aleks data tab, calculate the standard reaction free energy of the following chemical reaction: \\(\ce{c_{3}h_{8}(g) + 5o_{2}(g) -> 3co_{2}(g) + 4h_{2}o(l)}\\) round your answer to zero decimal places. \\(\square\\) kj \\(\square \times 10^{\square}\\)

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 ALEKS Data tab, typical values: $\Delta G_f^{\circ}(\text{C}_3\text{H}_8(g))=-23.4\ \text{kJ/mol}$, $\Delta G_f^{\circ}(\text{O}_2(g)) = 0\ \text{kJ/mol}$ (element in standard state), $\Delta G_f^{\circ}(\text{CO}_2(g))=-394.4\ \text{kJ/mol}$, $\Delta G_f^{\circ}(\text{H}_2\text{O}(l))=-237.1\ \text{kJ/mol}$)

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

For products: $3\ \text{mol of CO}_2$ and $4\ \text{mol of H}_2\text{O}(l)$
$\sum n\Delta G_f^{\circ}(\text{products})=3\times(-394.4)+4\times(-237.1)$
$= - 1183.2-948.4=-2131.6\ \text{kJ}$

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

For reactants: $1\ \text{mol of C}_3\text{H}_8(g)$ and $5\ \text{mol of O}_2(g)$
$\sum m\Delta G_f^{\circ}(\text{reactants})=1\times(-23.4)+5\times(0)$
$=-23.4 + 0=-23.4\ \text{kJ}$

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

$\Delta G^{\circ}=\sum n\Delta G_f^{\circ}(\text{products})-\sum m\Delta G_f^{\circ}(\text{reactants})$
$=-2131.6-(-23.4)$
$=-2131.6 + 23.4=-2108.2\ \text{kJ}$ (Wait, no, wait, correction: Wait, the formula is products - reactants, so it's $(-2131.6)-(-23.4)$? No, wait, reactants sum is $-23.4$, so products sum ($-2131.6$) minus reactants sum ($-23.4$) is $-2131.6+23.4=-2108.2$? Wait, no, let's re - calculate:

Wait, correct calculation:

Products:
$3\times\Delta G_f^{\circ}(\text{CO}_2) + 4\times\Delta G_f^{\circ}(\text{H}_2\text{O}(l))$
$=3\times(-394.4)+4\times(-237.1)$
$=-1183.2-948.4=-2131.6\ \text{kJ}$

Reactants:
$1\times\Delta G_f^{\circ}(\text{C}_3\text{H}_8) + 5\times\Delta G_f^{\circ}(\text{O}_2)$
$=1\times(-23.4)+5\times0=-23.4\ \text{kJ}$

Now, $\Delta G^{\circ}=\text{products - reactants}=-2131.6-(-23.4)=-2131.6 + 23.4=-2108.2\ \text{kJ}$? Wait, no, that can't be right. Wait, no, the formula is (sum of products' $\Delta G_f$) minus (sum of reactants' $\Delta G_f$). So if reactants sum is $-23.4$, then it's $(-2131.6)-(-23.4)=-2131.6 + 23.4=-2108.2$? But wait, actually, the correct typical value for this reaction (combustion of propane) has $\Delta G^{\circ}\approx - 2108\ \text{kJ}$ (when rounded to zero decimal places, - 2108? Wait, no, let's check with correct values. Wait, maybe I used wrong $\Delta G_f^{\circ}$ for $\text{C}_3\text{H}_8$. Let's use $\Delta G_f^{\circ}(\text{C}_3\text{H}_8(g))=-23.4\ \text{kJ/mol}$ (correct), $\text{O}_2$ is 0, $\text{CO}_2$ is - 394.4, $\text{H}_2\text{O}(l)$ is - 237.1.

Wait, another way: Let's recalculate products:

$3\times(-394.4)=-1183.2$

$4\times(-237.1)=-948.4$

Sum of products: $-1183.2-948.4=-2131.6$

Sum of reactants: $1\times(-23.4)+5\times0=-23.4$

$\Delta G^{\circ}=-2131.6-(-23.4)=-2131.6 + 23.4=-2108.2\ \text{kJ}$. Rounding to zero decimal places, we get - 2108? Wait, no, maybe the $\Delta G_f^{\circ}$ for $\text{C}_3\text{H}_8$ is different? Wait, actually, the correct standard free energy of formation for propane ($\text{C}_3\text{H}_8$) is - 23.4 kJ/mol, oxygen is 0, carbon dioxide is - 394.4 kJ/mol, liquid water is - 237.1 kJ/mol. So the calculation is correct. So $\Delta G^{\circ}=-2108.2\ \text{kJ}$, rounded to zero decimal places is - 2108? Wait, no, - 2108.2 rounded to zero decimal places is - 2108? Wait, no, - 2108.2, the first decimal is 2, so we round down? Wait, no, rounding to zero decimal places: look at the first decimal digit, which is 2, so we keep the integer part as - 2108? Wait, but maybe I made a mistake in the $\Delta G_f^{\circ}$ values. Let's check with the correct formula again.

Wait, the reaction is $\text{C}_3\text{H}_8(g)+5\text{O}_2(g)
ightarrow3\text{CO}_2(g)+4\text{H}_2\text{O}(l)$

So:

$\Delta G^{\circ}= [3\Delta G_f^{\circ}(\text{CO}_2) + 4\Delta G_f^{\circ}(\text{H}_2\text{O}(l))]-[ \Delta G_f^{\circ}(\text{C}_3\text{H}_8) + 5\Delta G_f^{\circ}(\text{O}_2) ]$

Plugging in the values:

$\Delta G_f^{\circ}(\text{CO}_2)=-394.4\ \text{kJ/mol}$, $\Delta G_f^{\circ}(\text{H}_2\text{O}(l))=-237.1\ \text{kJ/mol}$, $\Delta G_f^{\circ}(\text{C}_3\text{H}_8)=-23.4\ \text{kJ/mol}$, $\Delta G_f^{\circ}(\text{O}_2)=0\ \text{kJ/mol}$

So:

$3\times(-394.4)=-1183.2$

$4\times(-237.1)=-948.4$

Sum of products: $-1183.2-948.4=-2131.6$

Sum of reactants: $-23.4 + 5\times0=-23.4$

$\Delta G^{\circ}=-2131.6-(-23.4)=-2131.6 + 23.4=-2108.2\ \text{kJ}$

Rounding to zero decimal places, we get - 2108? Wait, no, - 2108.2, the decimal part is 0.2, so when rounding to zero decimal places, we look at the tenths place (2), which is less than 5, so we round down, so - 2108. But wait, maybe the actual value from ALEKS data is slightly different, but following the calculation with typical values, the result is - 2108 kJ (rounded to zero decimal places).

Answer:

-2108