91Ó°ÊÓ

When performing chemical calculations, understanding how to determine the number of moles is crucial. The mole is a fundamental unit in chemistry that allows us to translate between grams and individual molecules or atoms using the substance's molar mass. In this exercise, you are given 3.0 grams of carbon and need to convert this mass into moles, as the reaction's enthalpy is calculated using moles.

To find the moles of carbon, you take the given mass (3.0 g) and divide it by the molar mass of carbon, which is 12.01 g/mol. This calculation is expressed as:

\[ \text{moles of carbon} = \frac{3.0 \text{ g}}{12.01 \text{ g/mol}} \]

Plugging in the values, you get 0.25 moles of carbon. This tells us that in the reaction being analyzed, 0.25 moles of carbon will participate, and it is vital for calculating the exact enthalpy change for the given amount of carbon.

The enthalpy change of a reaction, symbolized as \( \Delta H \), reflects the heat that is either absorbed or released during a chemical reaction at constant pressure. It provides insight into the energy changes within a reaction. A negative \( \Delta H \) signifies an exothermic reaction, meaning the reaction releases heat. Conversely, a positive \( \Delta H \) indicates an endothermic process, absorbing heat.

For reaction (4), the enthalpy change needs to be determined using reactions (1), (2), and (3). Each of these reactions has an associated \( \Delta H \), showing how much energy is gained or lost. The task is to combine these reactions to find the overall enthalpy change for the formation of \( \mathrm{PbCO}_3(s) \) by combining them appropriately:

- Reaction (1): \( \Delta H_{1} = -219 \text{ kJ} \)
- Reaction (2): \( \Delta H_{2} = -394 \text{ kJ} \)
- Reaction (3): \( \Delta H_{3} = 86 \text{ kJ} \)

The enthalpy change for reaction (4) is then calculated using:

\[ \Delta H_{4} = \Delta H_{1} + \Delta H_{2} - \Delta H_{3} \]

Substituting in the given values results in \(-527 \text{ kJ/mol} \). This value must be adjusted to reflect the actual moles of carbon (0.25) we calculated, leading to an enthalpy change of \(-131.75 \text{ kJ}\) for the complete reaction of 3.0 g of carbon.

Hess's Law is a key principle in thermodynamics, stating that the total enthalpy change for a chemical reaction is the same, regardless of the pathway taken, when the initial and final conditions are the same. This is particularly useful in constructing enthalpy cycles or combining multiple reactions to determine the enthalpy change for a different reaction.

In this exercise, Hess's Law allows you to determine the enthalpy change of reaction (4) by combining the known enthalpy changes of reactions (1), (2), and (3). By rearranging and summing these reactions, one can "cancel out" the intermediate steps and focus on the overall transformation from reactants to products.

The practical application here is to use this law to deduce an unknown \( \Delta H \) from known ones. By summing the changes around the cycle (taking into account the signs for absorption and release of energy), you're able to accurately predict the enthalpy change for the complete process as seen in the given problem. This methodological power underlines why Hess's Law is a foundational tool in chemical thermodynamics.