Question

question 4 (1 point)
when 6.000 moles of glucose are combusted, 16 815 kj of energy is produced.
which of the following chemical equations represent this process?
6c₆h₁₂o₆(s) + 36o₂(g) → 36co₂(g) + 36h₂o(ℓ) dh = -2802.15 kj
6c₆h₁₂o₆(s) + 36o₂(g) → 36co₂(g) + 36h₂o(ℓ) dh = 16815 kj
c₆h₁₂o₆(s) + 6o₂(g) → 6co₂(g) + 6h₂o(ℓ) dh = -16815 kj
c₆h₁₂o₆(s) + 6o₂(g) → 6co₂(g) + 6h₂o(ℓ) dh = -2802.15 kj
all equations are correct

Explanation:

Step 1: Calculate energy per mole of glucose

Given 6.000 moles of glucose produce 16815 kJ. So per mole, energy change $\Delta H = \frac{-16815\ \text{kJ}}{6\ \text{moles}} = -2802.15\ \text{kJ/mol}$ (negative because combustion releases energy).

Step 2: Analyze the combustion reaction of glucose

The balanced equation for 1 mole of glucose ($\ce{C6H12O6}$) combustion is $\ce{C6H12O6(s) + 6O2(g) -> 6CO2(g) + 6H2O(\ell)}$ with $\Delta H = -2802.15\ \text{kJ}$ (since it's exothermic, $\Delta H$ is negative).

Step 3: Evaluate each option

• First option: Uses 6 moles of glucose, but the $\Delta H$ here is for 6 moles? No, its $\Delta H$ is -2802.15 which is per 1 mole. So incorrect.
• Second option: $\Delta H$ is positive (endothermic), but combustion is exothermic (negative $\Delta H$). Incorrect.
• Third option: $\Delta H$ is -16815 which is for 6 moles, but the equation is for 1 mole. Incorrect.
• Fourth option: Equation is for 1 mole of glucose, $\Delta H = -2802.15\ \text{kJ}$ (matches our calculation). Correct.
• Fifth option: Not all are correct.

Answer:

$\ce{C6H12O6(s) + 6O2(g) -> 6CO2(g) + 6H2O(\ell)}$ with $\Delta H = -2802.15\ \text{kJ}$ (the fourth option: $\boldsymbol{\ce{C6H12O6(s) + 6O2(g) -> 6CO2(g) + 6H2O(\ell)}}$ $\boldsymbol{\Delta H = -2802.15\ \text{kJ}}$)