Question

If the standard entropies of \(\mathrm{CH}_{4}, \mathrm{O}_{2}, \mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) are \(186.2,205.3,213.6\) and \(69.96 \mathrm{JK}^{-1} \mathrm{~mol}^{-\mathrm{t}}\) respectively, then standard entropy change for the reaction: \(\mathrm{CH}_{4}(\mathrm{~g})+2 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) is: (a) \(-215.6 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\) (b) \(-243.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\) (c) \(-130.5 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\) (d) \(-85.6 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\)

Short answer

The standard entropy change is \(-243.3 \, \mathrm{JK}^{-1} \mathrm{mol}^{-1}\), which is option (b).

Step-by-step solution

Write down the reaction

The given chemical reaction is \( \mathrm{CH}_{4} (\mathrm{g}) + 2 \mathrm{O}_{2} (\mathrm{g}) \to \mathrm{CO}_{2} (\mathrm{g}) + 2 \mathrm{H}_{2} \mathrm{O} (\mathrm{l}) \). We need to calculate the standard entropy change \( \Delta S^\circ \) for this reaction.

Identify standard entropies

The standard entropies of the substances involved in the reaction are: \( S^\circ(\mathrm{CH}_{4}) = 186.2 \, \mathrm{JK}^{-1} \mathrm{mol}^{-1} \), \( S^\circ(\mathrm{O}_{2}) = 205.3 \, \mathrm{JK}^{-1} \mathrm{mol}^{-1} \), \( S^\circ(\mathrm{CO}_{2}) = 213.6 \, \mathrm{JK}^{-1} \mathrm{mol}^{-1} \), and \( S^\circ(\mathrm{H}_{2} \mathrm{O}) = 69.96 \, \mathrm{JK}^{-1} \mathrm{mol}^{-1} \).

Calculate total entropy of products

For the products, total entropy \( S^\circ_\text{products} = S^\circ(\mathrm{CO}_{2}) + 2 \times S^\circ(\mathrm{H}_{2} \mathrm{O}) \). Substitute the values: \( S^\circ_\text{products} = 213.6 + 2 \times 69.96 \).

Calculate the result for products

Perform the arithmetic: \( S^\circ_\text{products} = 213.6 + 139.92 = 353.52 \, \mathrm{JK}^{-1} \mathrm{mol}^{-1} \).

Calculate total entropy of reactants

For the reactants, total entropy \( S^\circ_\text{reactants} = S^\circ(\mathrm{CH}_{4}) + 2 \times S^\circ(\mathrm{O}_{2}) \). Substitute the values: \( S^\circ_\text{reactants} = 186.2 + 2 \times 205.3 \).

Calculate the result for reactants

Perform the arithmetic: \( S^\circ_\text{reactants} = 186.2 + 410.6 = 596.8 \, \mathrm{JK}^{-1} \mathrm{mol}^{-1} \).

Calculate standard entropy change

Using the formula for standard entropy change, \( \Delta S^\circ = S^\circ_\text{products} - S^\circ_\text{reactants} \). Substitute: \( \Delta S^\circ = 353.52 - 596.8 \).

Solve for standard entropy change

Perform the subtraction: \( \Delta S^\circ = -243.28 \, \mathrm{JK}^{-1} \mathrm{mol}^{-1} \). This rounds to \( -243.3 \, \mathrm{JK}^{-1} \mathrm{mol}^{-1} \).

Choose the correct option

The calculation shows that \( \Delta S^\circ = -243.3 \, \mathrm{JK}^{-1} \mathrm{mol}^{-1} \), which corresponds to option (b).

Key concepts

Thermodynamics

Thermodynamics is a branch of physics that deals with the relationships between heat and other forms of energy. In the context of chemical reactions, it helps us understand how energy is transferred and transformed within a system. To better grasp thermodynamics, think of it as a set of rules governing how energy flows in and out of a system.
Within thermodynamics, the concept of entropy is crucial. Entropy is a measure of the degree of randomness or disorder in a system. In simple terms, it helps predict how and why energy changes occur. Always remember that nature tends to move towards a state of higher entropy or disorder.
When considering chemical reactions, thermodynamics provides insights into how variables such as temperature, pressure, and volume affect the outcome of a reaction. It also enables the calculation of whether a reaction will occur spontaneously. Spontaneous reactions are those that happen naturally without needing additional energy input from outside the system.

Chemical Reactions

Chemical reactions involve the transformation of one or more substances into different substances. These transformations involve the rearrangement of atoms and the breaking and forming of chemical bonds.
In our specific example, the reaction involved is the combustion of methane (\( \mathrm{CH}_{4} \)) with oxygen (\( \mathrm{O}_{2} \)), producing carbon dioxide and water. This type of reaction is an exothermic reaction, which means it releases heat.
Understanding the reactants and products in a chemical reaction is fundamental. Reactants are the starting substances that undergo change, while products are the substances formed as a result of the reaction. Every reaction is subject to the conservation of mass and energy, meaning the total mass and energy of the reactants equal that of the products.
In preparing for entropy calculations, it's helpful to write down the balanced equation of the reaction with state symbols like (g) for gas and (l) for liquid, as they indicate the physical state of substances, which can affect their entropy values.

Entropy Calculations

The standard entropy change (\( \Delta S^\circ \)) for a reaction is calculated to determine the change in disorder as reactants transform into products. It involves using the concept of standard entropy (\( S^\circ \)), which is the entropy content of a substance at a standard state, typically at 1 atm pressure and 298 K (25°C).

To calculate \( \Delta S^\circ \), follow these steps:

• Determine the total entropy of the products. This is done by summing up the standard entropy of each product multiplied by its stoichiometric coefficient from the balanced equation.
• Next, calculate the total entropy of the reactants similarly by summing up their individual entropies.
• Finally, use the formula: \( \Delta S^\circ = S^\circ_\text{products} - S^\circ_\text{reactants} \).

The result tells us whether the system gained or lost entropy. A negative \( \Delta S^\circ \) like in this scenario indicates a decrease in disorder, suggesting the products are more structured than the reactants.

Gibbs Free Energy

Gibbs Free Energy, represented as \( G \), is a crucial concept in understanding the spontaneity of chemical reactions. It combines enthalpy (\( H \)), entropy (\( S \)), and temperature (\( T \)) to predict the feasibility of a reaction under constant pressure and temperature.
The formula is given by:
\[ \Delta G = \Delta H - T \Delta S \]
Where:
- \( \Delta G \) is the change in Gibbs Free Energy.
- \( \Delta H \) is the change in enthalpy.
- \( T \) is the temperature in Kelvin.
- \( \Delta S \) is the change in entropy.
A negative \( \Delta G \) value indicates that a reaction occurs spontaneously, meaning it can proceed without external energy. If \( \Delta G \) is positive, the reaction is non-spontaneous and requires energy input.
Thus, Gibbs Free Energy plays a pivotal role in determining whether a reaction is thermodynamically favorable. It integrates the effects of enthalpy and entropy, providing a comprehensive picture of the energy dynamics in a reaction.