Question

Calculate \(\Delta S^{\circ}\) values for the following reactions by using tabulated \(S^{\circ}\) values from Appendix \(C\). In each case explain the sign of \(\Delta S^{\circ}\). (a) \(\mathrm{N}_{2} \mathrm{H}_{4}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{NH}_{3}(g)\) (b) \(\mathrm{K}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{KO}_{2}(s)\) (c) \(\mathrm{Mg}(\mathrm{OH})_{2}(s)+2 \mathrm{HCl}(\mathrm{g}) \longrightarrow \mathrm{MgCl}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l)\) (d) \(\mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(g)\)

Short answer

The standard entropy change, \(\Delta S^{\circ}\), for the given reactions are: (a) For the reaction: \(\mathrm N_2\mathrm H_4(g) + \mathrm H_2(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\), \(\Delta S^{\circ} = 132.8 \mathrm{~J~K^{-1}mol^{-1}}\). The positive value indicates an increase in disorder, due to two reactant molecules producing two new product molecules. (b) For the reaction: \(\mathrm K(s) + \mathrm O_2(g) \longrightarrow \mathrm{KO}_2(s)\), \(\Delta S^{\circ} = -148.5 \mathrm{~J~K^{-1}mol^{-1}}\). The negative value signifies a decrease in the system's disorder because one gas and one solid molecule react to form another solid molecule. (c) For the reaction: \(\mathrm{Mg}(\mathrm{OH})_{2}(s) + 2 \mathrm{HCl}(g) \longrightarrow \mathrm{MgCl}_{2}(s) + 2 \mathrm{H}_{2} \mathrm{O}(l)\), \(\Delta S^{\circ} = -207.5 \mathrm{~J~K^{-1}mol^{-1}}\). The negative value indicates a decrease in disorder because two gas molecules react with a solid, forming an additional solid and two liquid molecules. (d) For the reaction: \(\mathrm{CO}(g) + 2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{CH}_{3}\mathrm{OH}(g)\), \(\Delta S^{\circ} = -205.5 \mathrm{~J~K^{-1}mol^{-1}}\). The negative value suggests a decrease in disorder as three gas molecules react to form only one gas molecule.

Step-by-step solution

Find the standard molar entropy values from Appendix C

We can find the \(S^{\circ}\) values for the substances involved in the reaction from Appendix C: - \(S^{\circ}(\mathrm{N}_{2} \mathrm{H}_{4}(g)) = 121.6 \mathrm{~J~K^{-1}mol^{-1}}\) - \(S^{\circ}(\mathrm{H}_{2}(g)) = 130.6 \mathrm{~J~K^{-1}mol^{-1}}\) - \(S^{\circ}(\mathrm{NH}_{3}(g)) = 192.5 \mathrm{~J~K^{-1}mol^{-1}}\)

Calculate the standard entropy change for reaction (a)

Using the equation for the standard entropy change and the standard molar entropy values, we have: \[ \Delta S^{\circ} = 2S^{\circ}(\mathrm{NH}_{3}(g)) - [S^{\circ}(\mathrm{N}_{2} \mathrm{H}_{4}(g)) + S^{\circ}(\mathrm{H}_{2}(g))] \] \[ \Delta S^{\circ} = 2(192.5) - [121.6 + 130.6] \] \[ \Delta S^{\circ} = 385 - 252.2 = 132.8 \mathrm{~J~K^{-1}mol^{-1}} \] The standard entropy change for reaction (a) is positive, which means that the disorder of the system increases. This can be ascribed to the fact that two molecules of the reactant side produce two new molecules on the product side. ##Step 2: Calculate \(\Delta S^{\circ}\) for reaction (b)## (b) \(\mathrm{K}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{KO}_{2}(s)\)

Find the standard molar entropy values from Appendix C

We can find the \(S^{\circ}\) values for the substances involved in the reaction from Appendix C: - \(S^{\circ}(\mathrm{K}(s)) = 64.4 \mathrm{~J~K^{-1}mol^{-1}}\) - \(S^{\circ}(\mathrm{O}_{2}(g)) = 205.1 \mathrm{~J~K^{-1}mol^{-1}}\) - \(S^{\circ}(\mathrm{KO}_{2}(s)) = 121.0 \mathrm{~J~K^{-1}mol^{-1}}\)

Calculate the standard entropy change for reaction (b)

Using the equation for the standard entropy change and the standard molar entropy values, we have: \[ \Delta S^{\circ} = S^{\circ}(\mathrm{KO}_{2}(s)) - [S^{\circ}(\mathrm{K}(s)) + S^{\circ}(\mathrm{O}_{2}(g))] \] \[ \Delta S^{\circ} = 121 - [64.4 + 205.1] \] \[ \Delta S^{\circ} = 121 - 269.5 = -148.5 \mathrm{~J~K^{-1}mol^{-1}} \] The standard entropy change for reaction (b) is negative. This indicates a decrease in the disorder of the system, which is due to the fact that one gas and one solid molecule react, forming another solid molecule, which is more ordered than gaseous molecules. Thus, the reaction leads to a more ordered state. ##Step 3: Calculate \(\Delta S^{\circ}\) for reaction (c)## (c) \(\mathrm{Mg}(\mathrm{OH})_{2}(s)+2 \mathrm{HCl}(g) \longrightarrow \mathrm{MgCl}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l)\)

Find the standard molar entropy values from Appendix C

We can find the \(S^{\circ}\) values for the substances involved in the reaction from Appendix C: - \(S^{\circ}(\mathrm{Mg}(\mathrm{OH})_{2}(s)) = 63.4 \mathrm{~J~K^{-1}mol^{-1}}\) - \(S^{\circ}(\mathrm{HCl}(g)) = 186.9 \mathrm{~J~K^{-1}mol^{-1}}\) - \(S^{\circ}(\mathrm{MgCl}_{2}(s)) = 89.3 \mathrm{~J~K^{-1}mol^{-1}}\) - \(S^{\circ}(\mathrm{H}_{2} \mathrm{O}(l)) = 69.9 \mathrm{~J~K^{-1}mol^{-1}}\)

Calculate the standard entropy change for reaction (c)

Using the equation for the standard entropy change and the standard molar entropy values, we have: \[ \Delta S^{\circ} = S^{\circ}(\mathrm{MgCl}_{2}(s)) + 2S^{\circ}(\mathrm{H}_{2} \mathrm{O}(l)) - [S^{\circ}(\mathrm{Mg}(\mathrm{OH})_{2}(s)) + 2S^{\circ}(\mathrm{HCl}(g))] \] \[ \Delta S^{\circ} = 89.3 + 2(69.9) - [63.4 + 2(186.9)] \] \[ \Delta S^{\circ} = 89.3 + 139.8 - (63.4 + 373.8) = -207.5 \mathrm{~J~K^{-1}mol^{-1}} \] The standard entropy change for reaction (c) is negative, indicating a decrease in the disorder of the system. This is because two gas molecules react with one solid molecule, producing an additional solid and two liquid molecules. The transition from gases to liquids and solids results in a more ordered state. ##Step 4: Calculate \(\Delta S^{\circ}\) for reaction (d)## (d) \(\mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{CH}_{3}\mathrm{OH}(g)\)

Find the standard molar entropy values from Appendix C

We can find the \(S^{\circ}\) values for the substances involved in the reaction from Appendix C: - \(S^{\circ}(\mathrm{CO}(g)) = 197.6 \mathrm{~J~K^{-1}mol^{-1}}\) - \(S^{\circ}(\mathrm{H}_{2}(g)) = 130.6 \mathrm{~J~K^{-1}mol^{-1}}\) - \(S^{\circ}(\mathrm{CH}_{3}\mathrm{OH}(g)) = 253.3 \mathrm{~J~K^{-1}mol^{-1}}\)

Calculate the standard entropy change for reaction (d)

Using the equation for the standard entropy change and the standard molar entropy values, we have: \[ \Delta S^{\circ} = S^{\circ}(\mathrm{CH}_{3}\mathrm{OH}(g)) - [S^{\circ}(\mathrm{CO}(g)) + 2S^{\circ}(\mathrm{H}_{2}(g))] \] \[ \Delta S^{\circ} = 253.3 - [197.6 + 2(130.6)] \] \[ \Delta S^{\circ} = 253.3 - 458.8 = -205.5 \mathrm{~J~K^{-1}mol^{-1}} \] The standard entropy change for reaction (d) is negative, indicating a decrease in the disorder of the system. This is mainly due to the fact that three gas molecules react to form only one gas molecule. The reduction in the number of molecules leads to a decrease in the disorder within the system.

Key concepts

Thermodynamics

Thermodynamics is a branch of science that explores the relationship between heat, work, temperature, and energy in physical systems. It plays a crucial role in understanding how energy is transferred and transformed. This field is grounded in several laws, with the second law being particularly relevant to entropy and spontaneity in chemical reactions.

In thermodynamics, processes can be characterized by whether energy flows into or out of a system. This is often conceptualized using words like exothermic and endothermic. Exothermic processes release heat, whereas endothermic ones absorb heat. Entropy comes into the picture when we consider the randomness or disorder of a system.

Understanding thermodynamics helps explain why certain reactions occur and others do not. It also reveals the 'direction' energy might take during reactions. An important aspect to remember is that while energy tends to disperse, the specific path it takes is guided by both enthalpy and entropy changes.

Entropy Calculation

Entropy calculation is a key step in understanding the extent of disorder or randomness in a chemical system. The term 'entropy' itself is represented by the symbol 'S' and has units of Joules per kelvin per mole (J K⁻¹ mol⁻¹).

To calculate the entropy change (\(\Delta S^{\circ}\)), it is important to use standard molar entropy values (\(S^{\circ}\) ), which are typically found in appendices of chemistry textbooks or databases. The general formula used is:

• \[\Delta S^{\circ} = \sum S^{\circ}_{products} - \sum S^{\circ}_{reactants}\]

This formula is essential for determining the entropy change when substances react to form new compounds. Calculating entropy changes helps in predicting whether a reaction will result in an increase or decrease in disorder.

One critical outcome of entropy calculations is assessing reaction spontaneity. Generally, an increase in entropy (\(\Delta S^{\circ} > 0\)) suggests that a reaction is likely to be spontaneous under certain conditions. Keep in mind that spontaneity is also influenced by enthalpy and temperature changes, typically evaluated with the Gibbs free energy equation.

Chemical Reactions

Chemical reactions involve the transformation of substances as reactants turn into products. These reactions are not just about structural transformations but also involve changes in energy and entropy.

In a chemical reaction, reactant bonds break, and product bonds form. This process can lead to different energy and disorder levels in the system. For example, when gas molecules are involved, the phase and the number of molecules on either side of the equation play a significant role in dictating the system's entropy.

• Gas to solid or liquid transitions generally decrease entropy.
• An increase in the number of gas molecules typically increases entropy.

The nature of these transitions helps to determine the sign of \(\Delta S^{\circ}\). Positive values imply greater disorder, while negative values indicate the opposite.

Understanding these patterns is crucial when predicting how a reaction will behave. In essence, chemical reactions are dynamic processes tied closely with thermodynamic principles. Analyzing every component, from bond energies to entropy changes, provides insights into the forces driving these reactions.