Question

Calculate the enthalpy change for the combustion of cyclopropane at \(298 \mathrm{~K}\), if the enthalpy of formation \(\mathrm{CO}_{2}(\mathrm{~g}), \mathrm{H}_{2} \mathrm{O}(1)\) and propene \((\mathrm{g})\) are \(-393.5,-385.8\) and \(20.42 \mathrm{~kJ} \mathrm{~mol}^{-1}\) respectively. The enthalpy of isomerization of cyclopropane to propene is \(-33.0 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (a) \(1802 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (b) \(2091 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (c) \(2196 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (d) none

Short answer

(a) -1802 kJ/mol.

Step-by-step solution

Understanding the Reaction

The complete combustion of cyclopropane is represented as: \( C_3H_6 (g) + \frac{9}{2} O_2 (g) \rightarrow 3 CO_2 (g) + 3 H_2O (l) \). We need to find the enthalpy change for this reaction.

Apply Hess's Law

Use Hess's Law, which states that the total enthalpy change in a reaction is the sum of enthalpy changes for individual steps to calculate the desired enthalpy change.

Write Formation Equations

The enthalpy of formation for cyclopropane is not given directly. We use the isomerization of cyclopropane to propene: \( C_3H_6 (cyclopropane) \rightarrow C_3H_6 (propene) \) with \( \Delta H = -33.0 \mathrm{~kJ/mol} \). So, \( \Delta H_{formation}(cyclopropane) = \Delta H_{formation}(propene) + 33.0 \).

Calculate Enthalpy of Cyclopropane

Substitute the given value: \( \Delta H_{formation}(cyclopropane) = 20.42 + 33.0 = 53.42 \mathrm{~kJ/mol} \).

Calculate Total Enthalpy Change

Use the combustion equation: \( \Delta H_{combustion} = 3 \times \Delta H_{formation}(CO_2) + 3 \times \Delta H_{formation}(H_2O) - \Delta H_{formation}(cyclopropane) \). Substitute the values: \( 3(-393.5) + 3(-285.8) - 53.42 = -1802.52 \mathrm{~kJ/mol} \).

Compare with Options

Compare the calculated enthalpy change with the given options. Our calculation yields approximately \(-1802 \mathrm{~kJ/mol}\), which matches option (a).

Key concepts

Hess's Law

Hess's Law is an important concept in chemistry that simplifies the calculation of enthalpy changes in complex reactions. It states that the total enthalpy change in a reaction is equal to the sum of the enthalpy changes in the individual steps, regardless of the pathway taken. This means that we can break down a complicated reaction into simpler steps, calculate the enthalpy changes for these steps, and add them up to find the overall enthalpy change.

This approach is particularly helpful when direct measurement of a reaction's enthalpy isn't feasible. Instead, we can use known enthalpy changes of other reactions and apply Hess's Law. This law is based on the fact that enthalpy is a state function, meaning its value is determined only by the initial and final states of the system.

Summing up, Hess's Law allows us to compute the enthalpy change for any reaction by carefully choosing and combining other chemical equations, making it a powerful tool for understanding and predicting the energy changes in chemical processes.

Combustion Reaction

A combustion reaction is a chemical process that involves the rapid combination of a substance with oxygen to produce heat and light. This exothermic reaction is typically associated with burning, resulting in the formation of oxides, such as carbon dioxide and water, if the initial compound contains hydrogen and carbon.

In the case of cyclopropane, its combustion can be represented by the following balanced reaction:

• \[ C_3H_6 (g) + \frac{9}{2} O_2 (g) \rightarrow 3 CO_2 (g) + 3 H_2O (l) \]

Here, cyclopropane reacts with oxygen to form carbon dioxide and water.

Combustion reactions are not only essential in various industrial applications, like energy production, but also play a crucial role in natural processes, such as the carbon cycle. Understanding the enthalpy change associated with these reactions, which Hess's Law helps us calculate, is vital as it directly relates to the energy released or absorbed during the reaction.

Enthalpy of Formation

The enthalpy of formation is a measure of the heat change associated with forming one mole of a compound from its elements in their standard states. It's a fundamental concept because it provides useful information about the energy required to create compounds, helping us predict reaction outcomes.

Standard enthalpies of formation are usually denoted by \( \Delta H_{formation} \) and are measured under standard conditions (usually 1 atm pressure and 25°C). The enthalpy of formation for a pure element in its standard state is always zero because no formation from another element is necessary.

For instance, in the given problem, we know the enthalpy of formation for propene, carbon dioxide, and water. However, the enthalpy of formation for cyclopropane isn't given directly. Instead, it is calculated using this equation:

• \( \Delta H_{formation}(cyclopropane) = \Delta H_{formation}(propene) + 33.0 \)

This is due to the isomerization reaction of cyclopropane to propene.

By understanding enthalpies of formation, you can determine the energy changes for chemical reactions, such as combustion, using these standardized values and apply them in calculations employing Hess's Law.