Question

The heat evolved in the combustion of glucose, \(\mathrm{C}_{6} \mathrm{H}_{10} \mathrm{O}_{6}\) is \(-680 \mathrm{kcal} / \mathrm{mol}\). The mass of \(\mathrm{CO}_{2}\) produced, when \(170 \mathrm{kcal}\) of heat is evolved in the combustion of glucose is (a) \(45 \mathrm{~g}\) (b) \(66 \mathrm{~g}\) (c) \(11 \mathrm{~g}\) (d) \(44 \mathrm{~g}\)

Short answer

66 g of CO2 is produced when 170 kcal of heat is evolved in the combustion of glucose.

Step-by-step solution

Interpret the given heat of combustion

The heat of combustion of glucose, \(C_6H_{12}O_6\), is given as -680 kcal/mol. This means that for every 1 mole of glucose that is combusted, 680 kcal of heat is released.

Calculate moles of glucose corresponding to 170 kcal of heat

To find out how many moles of glucose correspond to 170 kcal of heat, set up a proportion where \(x\) is the number of moles corresponding to 170 kcal: \[ \frac{-680 \text{ kcal}}{1 \text{ mol}} = \frac{-170 \text{ kcal}}{x \text{ mol}} \] Solving for \(x\) gives: \[ x = \frac{-170 \text{ kcal}}{-680 \text{ kcal/mol}} \]

Solve for moles of glucose combusted

Calculate the moles of glucose for 170 kcal: \[ x = \frac{-170}{-680} = 0.25 \text{ mol} \]

Determine moles of CO2 produced per mole of glucose

The chemical equation for the combustion of glucose is: \[ C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O \] This indicates that for each mole of glucose combusted, 6 moles of CO2 are produced.

Calculate moles of CO2 produced from 0.25 mol of glucose

Since 1 mole of glucose produces 6 moles of CO2, 0.25 mole will produce: \[ 0.25 \text{ mol} \times 6 = 1.5 \text{ mol} \text{ of } CO2 \]

Find the mass of CO2 produced

The molar mass of CO2 is approximately 44 g/mol. To find the mass of CO2 produced from 1.5 moles, multiply the number of moles by the molar mass: \[ \text{Mass of } CO2 = 1.5 \text{ mol} \times 44 \text{ g/mol} \]

Solve for the mass of CO2 produced

Calculating the mass: \[ \text{Mass of } CO2 = 1.5 \text{ mol} \times 44 \text{ g/mol} = 66 \text{ g} \] Hence, the correct answer is (b) 66 g.

Key concepts

Heat of Combustion

Understanding the heat of combustion is essential when studying chemical reactions. It refers to the amount of heat released when a substance completely burns in the presence of oxygen. The heat of combustion is a negative value because it represents energy being released from the substance to the surroundings, indicating an exothermic process.

When looking at the example of glucose, \(C_6H_{12}O_6\), the given heat of combustion is -680 kcal/mol. This means that when one mole of glucose is burned, it emits 680 kilocalories of heat energy, making this an indicator of glucose's energy content. Learning to calculate the amount of heat involved when a known quantity of glucose is combusted is a practical application of this concept, which is pivotal in fields like nutrition, biochemistry, and energy production.

Stoichiometry

Stoichiometry is the part of chemistry that involves calculating the quantities of reactants and products in a chemical reaction. It relies on the law of conservation of mass, which states that in a chemical reaction, mass is neither created nor destroyed. To use stoichiometry, a balanced chemical equation is necessary because it shows the ratio of moles of reactants needed to moles of products formed.

In the chemical combustion of glucose, the stoichiometric coefficients show that one mole of glucose produces six moles of carbon dioxide (CO2). When working with stoichiometry problems, it is essential to clearly understand these ratios, as they guide the calculations for determining the amount of products formed from a given quantity of reactants and vice versa.

Chemical Thermodynamics

Chemical thermodynamics deals with the study of heat changes that accompany chemical reactions. It is an area of chemistry that links the concepts of heat transfer with the laws of thermodynamics, providing valuable insights into the spontaneity, energy changes, and equilibrium of reactions.

In the context of glucose combustion, thermodynamics helps determine the heat evolution as a measurable quantity of energy, which is a driving factor for the reaction's feasibility and extent. This relates directly to the exercise, where understanding the fundamental thermodynamic principles allows students to correlate the heat evolved with the moles of glucose burned and predict the outcome quantitatively.

Molar Mass Calculation

Molar mass calculation is vital for converting between the mass of a substance and the number of moles. The molar mass is the weight of one mole of a substance, typically expressed in grams per mole (g/mol). It is calculated using the atomic masses of the constituent elements and their proportions in a compound.

In our glucose combustion example, determining the molar mass of carbon dioxide, CO2, is a key step. Since CO2 has one carbon atom (approximately 12 g/mol) and two oxygen atoms (approximately 16 g/mol each), the molar mass corresponds to 44 g/mol. This allows us to convert moles of CO2 produced into grams, facilitating the connection between theoretical calculations and tangible, measurable quantities in a real-world setting.