Question

using the thermodynamic information in the aleks data tab, calculate the standard reaction free energy of the following chemical reaction:
2h₂(g) + o₂(g) → 2h₂o(g)
round your answer to zero decimal places.
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 standard free energy of formation values

From thermodynamic data (ALEKS Data tab, typically):
- $\Delta G_f^\circ(\text{H}_2(g)) = 0\ \text{kJ/mol}$ (elements in standard state have $\Delta G_f^\circ = 0$)
- $\Delta G_f^\circ(\text{O}_2(g)) = 0\ \text{kJ/mol}$ (same reason as above)
- $\Delta G_f^\circ(\text{H}_2\text{O}(g)) = -228.57\ \text{kJ/mol}$ (standard value for gaseous water)

Step3: Apply the formula to the reaction

For the reaction $2\text{H}_2(g) + \text{O}_2(g)
ightarrow 2\text{H}_2\text{O}(g)$:
- Sum of $\Delta G_f^\circ$ for products: $2 \times \Delta G_f^\circ(\text{H}_2\text{O}(g)) = 2 \times (-228.57\ \text{kJ/mol}) = -457.14\ \text{kJ/mol}$
- Sum of $\Delta G_f^\circ$ for reactants: $2 \times \Delta G_f^\circ(\text{H}_2(g)) + 1 \times \Delta G_f^\circ(\text{O}_2(g)) = 2 \times 0 + 1 \times 0 = 0\ \text{kJ/mol}$
- $\Delta G^\circ = (-457.14) - 0 = -457.14\ \text{kJ/mol}$ (for the reaction as written, with stoichiometry, so the total free energy change for the reaction is also -457.14 kJ when considering the moles, and rounding to zero decimal places gives -457 kJ? Wait, no, wait: Wait, the stoichiometry is 2 moles of H2O, so let's check again. Wait, the calculation is per mole of reaction? Wait, no, the formula is based on the stoichiometric coefficients. Let's recast:

The reaction is $2\text{H}_2(g) + \text{O}_2(g)
ightarrow 2\text{H}_2\text{O}(g)$. So the number of moles for products: 2 moles of H2O, reactants: 2 moles H2, 1 mole O2.

So $\Delta G^\circ = [2 \times \Delta G_f^\circ(\text{H}_2\text{O}(g))] - [2 \times \Delta G_f^\circ(\text{H}_2(g)) + 1 \times \Delta G_f^\circ(\text{O}_2(g))]$

Plugging in the values:

$\Delta G_f^\circ(\text{H}_2\text{O}(g)) = -228.57\ \text{kJ/mol}$, so 2 moles: $2 \times (-228.57) = -457.14\ \text{kJ}$ (since the units for $\Delta G_f^\circ$ are per mole, and the reaction produces 2 moles of H2O, so the total free energy change for the reaction is -457.14 kJ, and rounding to zero decimal places gives -457 kJ? Wait, no, wait: Wait, the standard value for $\Delta G_f^\circ$ of H2O(g) is -228.57 kJ/mol. So for 2 moles, it's 2*(-228.57) = -457.14 kJ. The reactants have $\Delta G_f^\circ$ of 0, so the total $\Delta G^\circ$ is -457.14 kJ, which rounds to -457 kJ? Wait, but maybe the ALEKS data has a slightly different value? Wait, sometimes the value is -228.6 kJ/mol, so 2*(-228.6) = -457.2, which rounds to -457 kJ. But let's confirm:

Wait, actually, the correct standard free energy of formation for H2O(g) is -228.572 kJ/mol (more precise). So 2*(-228.572) = -457.144 kJ. Rounding to zero decimal places: -457 kJ. But wait, maybe the problem expects using the value from ALEKS, which might be -228.6, so 2*(-228.6) = -457.2, rounded to -457? Wait, no, -457.144 rounded to zero decimal places is -457 (since the first decimal is 1, which is less than 5? Wait, no: -457.144, the decimal part is 0.144, so when rounding to zero decimal places, we look at the first decimal digit, which is 1, so we round down, keeping the integer part as -457. Wait, but wait, maybe I made a mistake in the sign? Wait, the reaction is forming water from hydrogen and oxygen, which is a spontaneous reaction, so $\Delta G^\circ$ should be negative, which matches.

Wait, but let's check again. The formula is $\Delta G^\circ = \sum
u \Delta G_f^\circ(\text{products}) - \sum
u \Delta G_f^\circ(\text{reactants})$, where $
u$ is the stoichiometric coefficient (positive for products, negative for reactants? Wait, no: the formula is (sum of products' $\Delta G_f^\circ$ multiplied by their coefficients) minus (sum of reactants' $\Delta G_f^\circ$ multiplied by their coefficients). So for products: 2 moles of H2O, so 2*$\Delta G_f^\circ$(H2O). For reactants: 2 moles of H2 and 1 mole of O2, so 2*$\Delta G_f^\circ$(H2) + 1*$\Delta G_f^\circ$(O2). Since H2 and O2 are elements in standard state, their $\Delta G_f^\circ$ is 0. So $\Delta G^\circ = (2*(-228.57)) - (2*0 + 1*0) = -457.14$ kJ. Rounding to zero decimal places: -457 kJ.

But wait, maybe the ALEKS data has $\Delta G_f^\circ$ of H2O(g) as -228.6 kJ/mol, so 2*(-228.6) = -457.2, which rounds to -457 kJ (since the first decimal is 2, which is less than 5? Wait, no: -457.2, the decimal is 0.2, so when rounding to zero decimal places, we look at the tenths place: 2 < 5, so we round down, getting -457. Alternatively, if the value is -228.57, then 2*(-228.57) = -457.14, which rounds to -457 (since 0.14 < 0.5). So the answer should be -457 kJ? Wait, but let's check with the exact calculation.

Wait, maybe I messed up the stoichiometry. Let's re-express the reaction:

Reactants: 2 H2(g) + 1 O2(g)

Products: 2 H2O(g)

$\Delta G^\circ = [2 \times \Delta G_f^\circ(\text{H}_2\text{O}(g))] - [2 \times \Delta G_f^\circ(\text{H}_2(g)) + 1 \times \Delta G_f^\circ(\text{O}_2(g))]$

$\Delta G_f^\circ(\text{H}_2(g)) = 0$, $\Delta G_f^\circ(\text{O}_2(g)) = 0$, $\Delta G_f^\circ(\text{H}_2\text{O}(g)) = -228.57\ \text{kJ/mol}$

So:

$\Delta G^\circ = 2*(-228.57) - (2*0 + 1*0) = -457.14\ \text{kJ}$

Rounding to zero decimal places: -457 kJ.

Answer:

-457