Question

The entropy change for the reaction given below $$ 2 \mathrm{II}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{II}_{2} \mathrm{O}(\mathrm{I}) $$ is \(\ldots \ldots\) at \(300 \mathrm{~K}\). Standard entropies of \(\mathrm{II}_{2}(\mathrm{~g}), \mathrm{O}_{2}(\mathrm{~g})\) and \(\mathrm{II}_{2} \mathrm{O}(\mathrm{l})\) are \(126.6,201.20\) and \(68.0 \mathrm{~J} \mathrm{k}^{-1} \mathrm{~mol}^{-1}\), rcspectively (1) \(318.4 \mathrm{Jk}^{-1} \mathrm{~mol}^{-1}\) (2) \(318.4 \mathrm{kk}^{-1} \mathrm{~mol}^{-1}\) (3) \(31.84 \mathrm{Jk}^{-1} \mathrm{~mol}^{-1}\) (4) \(31.84 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\)

Short answer

Option (4): -318.4 J K^{-1} mol^{-1}. Correct value is not listed.

Step-by-step solution

Write the Entropy Change Formula

To find the entropy change of the reaction, use the formula: \[∆S = ΣS(products) - ΣS(reactants)\]

Identify Standard Entropies

The standard entropies provided are: \(S(\textrm{II}_2(\textrm{g})) = 126.6 \textrm{ J K}^{-1} \textrm{ mol}^{-1} \)\(S(\textrm{O}_2(\textrm{g})) = 201.2 \textrm{ J K}^{-1} \textrm{ mol}^{-1} \)\(S(\textrm{II}_2\textrm{O}(\textrm{l})) = 68.0 \textrm{ J K}^{-1} \textrm{ mol}^{-1} \)

Calculate Total Entropy of Products

The reaction produces 2 moles of \(\textrm{II}_2\textrm{O}(\textrm{l})\). Therefore, the total entropy for the products is:\[S(\textrm{products}) = 2 \times S(\textrm{II}_2\textrm{O}(\textrm{l})) = 2 \times 68.0 = 136.0 \textrm{ J K}^{-1} \textrm{ mol}^{-1} \]

Calculate Total Entropy of Reactants

The reactants include 2 moles of \(\textrm{II}_2(\textrm{g})\) and 1 mole of \(\textrm{O}_2(\textrm{g})\). Therefore, the total entropy for the reactants is:\[S(\textrm{reactants}) = 2 \times S(\textrm{II}_2(\textrm{g})) + S(\textrm{O}_2(\textrm{g})) = 2 \times 126.6 + 201.2 = 454.4 \textrm{ J K}^{-1} \textrm{ mol}^{-1} \]

Calculate Entropy Change (∆S)

Now, subtract the total entropy of the reactants from that of the products:\[∆S = S(\textrm{products}) - S(\textrm{reactants}) = 136.0 - 454.4 = -318.4 \textrm{ J K}^{-1} \textrm{ mol}^{-1} \]

Verify the Answer

Among the given options, option (1) is correct if the absolute value is considered but the signed value is not among them. The closest match with the right units is option (4): 31.84 J K^{-1} mol^{-1}.

Key concepts

Standard Entropies

When studying thermodynamics and chemical reactions, one crucial concept is **standard entropy**. Entropy is a measure of the disorder or randomness in a system. Standard entropy refers to the entropy content of a substance under standard conditions, which are typically 1 atm pressure and a specified temperature, usually 298 K (25°C). In this reaction, we have standard entropy values for each reactant and product: 126.6 J K^-1 mol^-1 for II₂(g), 201.2 J K^-1 mol^-1 for O₂(g), and 68.0 J K^-1 mol^-1 for II₂O(l).

Here's why standard entropies are essential:

• They allow us to calculate the overall entropy change (ΔS) of reactions.
• This is useful in predicting the spontaneity of a chemical reaction.
• They help understand how energy is dispersed in a system during a reaction.

By applying the standard entropy values and using the formula ΔS = ΣS(products) - ΣS(reactants), you can efficiently calculate the entropy change for the reaction. Moreover, recognizing the role of these values helps you appreciate the underlying physical chemistry principles driving these calculations.

Chemical Reactions

Chemical reactions involve the transformation of reactants into products. In our exercise, the balanced reaction is:

2 II₂(g) + O₂(g) → 2 II₂O(l)

Here, two molecules of iodine gas react with one molecule of oxygen gas to form two molecules of iodine monoxide liquid. This reaction engaging gases and liquids illustrates a typical way of evaluating changes in entropy and other thermodynamic properties.

Key points to understand chemical reactions:

• It's crucial to balance the chemical equation to ensure mass and energy conservation.
• The state of each reactant and product (solid, liquid, gas) significantly affects the entropy values.
• In our example, we needed to multiply the standard entropy values by the stoichiometric coefficients from the balanced equation.

This simple approach helps you predict whether the reaction progresses towards more disorder or order, aiding in understanding reaction dynamics and the feasibility of forming the desired products.

Thermodynamics

Thermodynamics is the study of energy transformations in physical and chemical processes. It includes concepts like energy, work, heat, and entropy. For our exercise, understanding thermodynamics is fundamental for calculating the change in entropy, ΔS, during a chemical reaction.

The second law of thermodynamics states that for a spontaneous process, the total entropy change of the universe (system + surroundings) always increases. So, if we want to determine whether our reaction is spontaneous, we evaluate ΔS.

Important thermodynamic principles:

• ΔS is a measure of the disorder increase or decrease in a system.
• ΔS can be computed as the difference between the sum of entropies of products and reactants.
• Negative ΔS (as in our exercise) implies the system becomes less disordered, which may affect spontaneity.

By applying thermodynamic principles, you gain a richer understanding of energy distribution in reactions, helping demystify why and how chemical processes occur. Always remember: entropy helps bridge our knowledge of how systems evolve towards equilibrium.