Question

Glucose is the main fuel metabolized in animal cells: \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}+6 \mathrm{O}_{2} \rightarrow 6 \mathrm{CO}_{2}+6 \mathrm{H}_{2} \mathrm{O} \Delta H=-2,799 \mathrm{~kJ}\) How much energy is given off when \(100.0 \mathrm{~g}\) of \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) react?

Short answer

The reaction releases approximately -1554.4 kJ of energy.

Step-by-step solution

Calculate the Molar Mass of Glucose

The molecular formula for glucose is \( \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} \). We need to find the molar mass by adding up the atomic masses: \( 6 \times 12.01 \) for carbon, \( 12 \times 1.008 \) for hydrogen, and \( 6 \times 16.00 \) for oxygen. This calculation gives: \[ 6 \times 12.01 + 12 \times 1.008 + 6 \times 16.00 = 180.16 \text{ g/mol} \].

Calculate Moles of Glucose

To find out how many moles are in \( 100.0 \text{ g} \) of glucose, use the formula: \( \text{moles} = \frac{\text{mass}}{\text{molar mass}} \). Substitute the given values: \[ \text{moles} = \frac{100.0 \text{ g}}{180.16 \text{ g/mol}} \approx 0.555 \text{ mol} \].

Calculate Energy Released

According to the balanced reaction: \( \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} + 6 \mathrm{O}_{2} \rightarrow 6 \mathrm{CO}_{2} + 6 \mathrm{H}_{2} \mathrm{O} \Delta H = -2799 \text{ kJ} \). This means \( -2799 \text{ kJ} \) is released for 1 mole of glucose. For \( 0.555 \text{ mol} \), the energy is \( \Delta H = 0.555 \times -2799 \text{ kJ/mol} \approx -1554.4 \text{ kJ} \).

Key concepts

Understanding Glucose Metabolism

Glucose metabolism is a fundamental biochemical process that occurs in nearly every cell of living organisms, especially animal cells. This process converts glucose, a simple sugar, into energy through a series of reactions, ensuring that cells have the necessary power to perform vital functions. The primary pathway for glucose metabolism is glycolysis, where glucose molecules are broken down into pyruvate, yielding ATP (adenosine triphosphate), the energy currency of cells.

• The complete aerobic metabolism of glucose can be represented by the equation: \( \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} + 6 \mathrm{O}_{2} \rightarrow 6 \mathrm{CO}_{2} + 6 \mathrm{H}_{2} \mathrm{O} \).
• This reaction not only involves the breakdown of glucose but also the release of energy.
• The term \( \Delta H = -2799 \text{ kJ} \) represents the change in enthalpy, indicating that the process is exothermic and releases heat.

Understanding glucose metabolism is essential because it is intricately linked to how organisms produce energy efficiently and maintain homeostasis.

Molar Mass Calculation Made Simple

Calculating the molar mass is a crucial step in many chemistry problems. It involves determining the mass of one mole of a substance based on its chemical formula. For glucose, which has the molecular formula \( \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} \), we find its molar mass by adding the atomic masses of its constituent elements.

• Carbon (C) has an atomic mass of approximately 12.01 g/mol. With 6 carbon atoms in glucose, the contribution is \( 6 \times 12.01 \).
• Hydrogen (H) has an atomic mass of 1.008 g/mol, and with 12 hydrogen atoms, it contributes \( 12 \times 1.008 \).
• Oxygen (O) has an atomic mass of 16.00 g/mol, contributing \( 6 \times 16.00 \).

Adding these values gives the molar mass:\[ 6 \times 12.01 + 12 \times 1.008 + 6 \times 16.00 = 180.16 \text{ g/mol} \].
This calculation ensures that we correctly use this number in further chemisry calculations like determining moles or energy changes.

Exploring Enthalpy Change in Reactions

Enthalpy change is a key concept in thermochemistry. It indicates whether a reaction absorbs or releases heat. For glucose oxidation, the enthalpy change \( \Delta H = -2799 \text{ kJ/mol} \) tells us a substantial amount of energy is released when one mole of glucose reacts with oxygen.

• The negative sign in \( \Delta H \) denotes that the reaction is exothermic.
• This means energy is transferred from the system to the surroundings, making it an energy-releasing process.
• In practice, knowing the enthalpy change helps predict the energetic feasibility of reactions and their environmental impact.
• In this case, it explains how glucose metabolism efficiently provides energy for cellular activities.

In a real-world context, understanding enthalpy change is vital for designing industrial chemical processes and for comprehending metabolic pathways because it impacts energy conservation and transformation crucial for life.