Question

formation of glucose
gibbs free energy
the formation of glucose from carbon dioxide and
water occurs according to the equation below.

6co₂(g) + 6h₂o(g) → c₆h₁₂o₆(s) + 6o₂(g)

δgᵣₓₙ =?

substance | δgᵣ° (kj/mol)
c₆h₁₂o₆(g) | -2868
co₂(g) | -394
o₂(g) | 0
h₂o(g) | -229

using the equation and data table, calculate the
gibbs free energy for the glucose reaction.
-2245 kj
-870 kj
2245 kj
870 kj

Explanation:

Step1: Recall the formula for Gibbs Free Energy of reaction

The formula for the Gibbs Free Energy change of a reaction ($\Delta G_{rxn}$) is given by the sum of the Gibbs Free Energy of formation ($\Delta G_f^{\circ}$) of the products minus the sum of the Gibbs Free Energy of formation of the reactants, multiplied by their respective stoichiometric coefficients. Mathematically, it is:
$$\Delta G_{rxn} = \sum (n \times \Delta G_f^{\circ}_{\text{products}}) - \sum (m \times \Delta G_f^{\circ}_{\text{reactants}})$$
where $n$ and $m$ are the stoichiometric coefficients of the products and reactants respectively.

Step2: Identify the products and reactants with their stoichiometric coefficients

For the reaction: $6\text{CO}_2(g) + 6\text{H}_2\text{O}(g)
ightarrow \text{C}_6\text{H}_{12}\text{O}_6(s) + 6\text{O}_2(g)$
- **Products**: $\text{C}_6\text{H}_{12}\text{O}_6(s)$ (stoichiometric coefficient $n = 1$) and $\text{O}_2(g)$ (stoichiometric coefficient $n = 6$)
- **Reactants**: $\text{CO}_2(g)$ (stoichiometric coefficient $m = 6$) and $\text{H}_2\text{O}(g)$ (stoichiometric coefficient $m = 6$)

Step3: Find the $\Delta G_f^{\circ}$ values for each substance

From the data table:
- $\Delta G_f^{\circ}(\text{C}_6\text{H}_{12}\text{O}_6(s)) = -2868$ kJ/mol (Wait, the table has $\text{C}_6\text{H}_{12}\text{O}_6(g)$ but the product is solid? Wait, maybe it's a typo, but let's check the table again. Wait the table says $\text{C}_6\text{H}_{12}\text{O}_6(g)$ with $\Delta G_f^{\circ} = -2868$ kJ/mol. Maybe it's a typo and should be solid, but we'll use the given values.
- $\Delta G_f^{\circ}(\text{O}_2(g)) = 0$ kJ/mol
- $\Delta G_f^{\circ}(\text{CO}_2(g)) = -394$ kJ/mol
- $\Delta G_f^{\circ}(\text{H}_2\text{O}(g)) = -229$ kJ/mol

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

For products:
- $\text{C}_6\text{H}_{12}\text{O}_6(s)$ (or gas as per table): $n = 1$, so $1 \times (-2868)$ kJ/mol
- $\text{O}_2(g)$: $n = 6$, so $6 \times 0$ kJ/mol
Sum of products: $1\times(-2868) + 6\times0 = -2868$ kJ

Step5: Calculate the sum of $\Delta G_f^{\circ}$ for reactants

For reactants:
- $\text{CO}_2(g)$: $m = 6$, so $6 \times (-394)$ kJ/mol
- $\text{H}_2\text{O}(g)$: $m = 6$, so $6 \times (-229)$ kJ/mol
First, calculate for $\text{CO}_2$: $6\times(-394) = -2364$ kJ
For $\text{H}_2\text{O}$: $6\times(-229) = -1374$ kJ
Sum of reactants: $-2364 + (-1374) = -3738$ kJ

Step6: Calculate $\Delta G_{rxn}$

Using the formula:
$$\Delta G_{rxn} = (\text{Sum of products}) - (\text{Sum of reactants})$$
Substitute the values:
$$\Delta G_{rxn} = (-2868) - (-3738)$$
$$\Delta G_{rxn} = -2868 + 3738$$
$$\Delta G_{rxn} = 870$$ Wait, that can't be right. Wait, maybe I mixed up products and reactants. Wait the formula is $\sum$ (products) - $\sum$ (reactants). Wait let's re-express:

Wait the reaction is $6\text{CO}_2 + 6\text{H}_2\text{O}
ightarrow \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2$

So products are $\text{C}_6\text{H}_{12}\text{O}_6$ (1 mol) and $\text{O}_2$ (6 mol)
Reactants are $\text{CO}_2$ (6 mol) and $\text{H}_2\text{O}$ (6 mol)

So $\sum \Delta G_f^{\circ}_{\text{products}} = [\Delta G_f^{\circ}(\text{C}_6\text{H}_{12}\text{O}_6) + 6\times\Delta G_f^{\circ}(\text{O}_2)]$
$\sum \Delta G_f^{\circ}_{\text{reactants}} = [6\times\Delta G_f^{\circ}(\text{CO}_2) + 6\times\Delta G_f^{\circ}(\text{H}_2\text{O})]$

Then $\Delta G_{rxn} = \sum \text{products} - \sum \text{reactants}$

Plugging in the values:

$\Delta G_f^{\circ}(\text{C}_6\text{H}_{12}\text{O}_6) = -2868$ (from table, even though it's gas, maybe the problem has a typo)
$\Delta G_f^{\circ}(\text{O}_2) = 0$
$\Delta G_f^{\circ}(\text{CO}_2) = -394$
$\Delta G_f^{\circ}(\text{H}_2\text{O}) = -229$

So:

Sum of products: $(-2868) + 6\times0 = -2868$

Sum of reactants: $6\times(-394) + 6\times(-229) = 6\times(-394 - 229) = 6\times(-623) = -3738$

Then $\Delta G_{rxn} = (-2868) - (-3738) = -2868 + 3738 = 870$? Wait but that's one of the options. Wait but let's check the reaction again. Wait the product is $\text{C}_6\text{H}_{12}\text{O}_6(s)$ but the table has $\text{C}_6\text{H}_{12}\text{O}_6(g)$. Maybe it's a mistake, but according to the given data, let's proceed.

Wait, maybe I made a mistake in the formula. Wait the formula is $\Delta G_{rxn} = \sum (n \Delta G_f^{\circ}_{\text{products}}) - \sum (m \Delta G_f^{\circ}_{\text{reactants}})$

So products: $\text{C}_6\text{H}_{12}\text{O}_6$ (1 mol) and $\text{O}_2$ (6 mol)
Reactants: $\text{CO}_2$ (6 mol) and $\text{H}_2\text{O}$ (6 mol)

So:

$\sum$ products: $1\times(-2868) + 6\times0 = -2868$

$\sum$ reactants: $6\times(-394) + 6\times(-229) = 6\times(-394 - 229) = 6\times(-623) = -3738$

Then $\Delta G_{rxn} = (-2868) - (-3738) = 870$ kJ? But wait the options include 870 kJ. Wait but let's check the reaction again. Wait the reaction is the formation of glucose, which is a photosynthesis-like reaction? Wait no, photosynthesis is endergonic, but the given reaction is $6\text{CO}_2 + 6\text{H}_2\text{O}
ightarrow \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2$. Wait the Gibbs free energy for this reaction in reality is positive (endergonic), but according to the calculation, it's 870 kJ? Wait but the options have 870 kJ as an option. Wait maybe the table has a typo, and the $\text{C}_6\text{H}_{12}\text{O}_6$ is solid with $\Delta G_f^{\circ} = -910$? No, the given table has -2868. Wait maybe I mixed up the formula. Wait no, the formula is correct. Let's recalculate:

Sum of products: $\Delta G_f^{\circ}(\text{C}_6\text{H}_{12}\text{O}_6) + 6\times\Delta G_f^{\circ}(\text{O}_2) = -2868 + 6\times0 = -2868$

Sum of reactants: $6\times\Delta G_f^{\circ}(\text{CO}_2) + 6\times\Delta G_f^{\circ}(\text{H}_2\text{O}) = 6\times(-394) + 6\times(-229) = -2364 - 1374 = -3738$

Then $\Delta G_{rxn} = (-2868) - (-3738) = -2868 + 3738 = 870$ kJ. So the answer should be 870 kJ.

Answer:

870 kJ (corresponding to the option "870 kJ")