Question

formation of glucose
balanced equation
most chemical reactions are considered reversible but are often only thermodynamically favorable in one direction. plants generate glucose from carbon dioxide and water.
the formula for glucose is c₆h₁₂o₆.
which of the following equations represents the balanced reaction in which glucose and oxygen are formed from carbon dioxide and water?
c₆h₁₂o₆(g) + 6o₂(g) → 6co₂(g) + 6h₂o(g)
3co₂(g) + 3h₂o(g) → c₆h₁₂o₆(g) + 3o₂(g)
6co₂(g) + 6h₂o(g) → c₆h₁₂o₆(g) + 6o₂(g)
co₂(g) + h₂o(g) → c₆h₁₂o₆(g) + o₂(g)

Explanation:

To determine the balanced equation for the formation of glucose ($\ce{C6H12O6}$) and oxygen from carbon dioxide ($\ce{CO2}$) and water ($\ce{H2O}$), we balance the number of each atom (C, H, O) on both sides of the equation.

1. Carbon (C) atoms: Glucose has 6 C atoms, so we need 6 $\ce{CO2}$ (since each $\ce{CO2}$ has 1 C).
2. Hydrogen (H) atoms: Glucose has 12 H atoms, so we need 6 $\ce{H2O}$ (since each $\ce{H2O}$ has 2 H: $6 \times 2 = 12$).
3. Oxygen (O) atoms:

• Left side (reactants): $6\ce{CO2}$ has $6 \times 2 = 12$ O; $6\ce{H2O}$ has $6 \times 1 = 6$ O. Total O: $12 + 6 = 18$.
• Right side (products): Glucose ($\ce{C6H12O6}$) has 6 O; oxygen ($\ce{O2}$) contributes the rest. Let $x$ be the moles of $\ce{O2}$. Then $6 + 2x = 18 \implies 2x = 12 \implies x = 6$. So we need 6 $\ce{O2}$.

Checking the options:

• First option: Reverses the reaction (glucose + $\ce{O2}$ → $\ce{CO2}$ + $\ce{H2O}$), which is combustion, not formation.
• Second option: Incorrect C and O balancing (3 $\ce{CO2}$ gives 3 C, but glucose has 6 C).
• Third option: $6\ce{CO2} + 6\ce{H2O}

ightarrow \ce{C6H12O6} + 6\ce{O2}$: Balances C (6), H (12), and O (18 on both sides).

• Fourth option: Unbalanced (1 C on left, 6 on right; 2 H on left, 12 on right; etc.).

Answer:

C. $6\ce{CO2(g)} + 6\ce{H2O(g)}
ightarrow \ce{C6H12O6(g)} + 6\ce{O2(g)}$