Use data from Appendix IIB to calculate ΔSrnn^∘ for each of the reactions. In each case, try to

Use data from Appendix IIB to calculate ΔSrnn^∘ for each of...

Use data from Appendix IIB to calculate $\Delta S_{\mathrm{rnn}}^{\circ}$ for each of the reactions. In each case, try to rationalize the sign of $\Delta S_{\mathrm{rxn}}^{\circ}$ a. $\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)$ b. $\mathrm{C}(s)+\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{CO}(g)+\mathrm{H}_{2}(g)$ c. $\mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g)$ d. $2 \mathrm{H}_{2} \mathrm{~S}(g)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{SO}_{2}(g)$

Use data from Appendix IIB to calculate $\Delta S_{\mathrm{rnn}}^{\circ}$ for each of the reactions. In each case, try to rationalize the sign of $\Delta S_{\mathrm{rxn}}^{\circ}$
a. $\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)$
b. $\mathrm{C}(s)+\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{CO}(g)+\mathrm{H}_{2}(g)$
c. $\mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g)$
d. $2 \mathrm{H}_{2} \mathrm{~S}(g)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{SO}_{2}(g)$

Key Concepts

Standard Molar Entropy

Standard molar entropy is a measure of the absolute entropy of a substance under standard conditions, typically given in units of joules per mole kelvin (J/mol·K). It represents the degree of randomness or disorder in a system and is tabulated for chemicals in their standard states. This concept is essential in thermodynamics for comparing the entropy values of different substances involved in reactions.

Standard Reaction Entropy

Standard reaction entropy is the change in entropy that occurs when a reaction takes place under standard conditions. It is calculated by subtracting the sum of the standard molar entropies of the reactants, each multiplied by its stoichiometric coefficient, from the sum of the standard molar entropies of the products similarly weighted. This calculation allows chemists to quantitatively assess how a reaction influences the overall disorder of a system.

Stoichiometry in Entropy Calculations

Stoichiometry plays a critical role in entropy calculations because the total entropy change of a reaction depends on the relative amounts of each substance participating. Multiplying the standard molar entropy of each reactant or product by its respective stoichiometric coefficient ensures that the calculated reaction entropy accurately reflects the contributions of all species present in the chemical equation.

Effects of Molecular Phases on Entropy

The physical state of a substance — whether it is a gas, liquid, or solid — greatly affects its entropy. Generally, gases have higher entropy compared to liquids and solids due to a greater freedom of movement and larger number of accessible microstates. Recognizing the phase of each reactant and product is important when evaluating the entropy change in a reaction.

Rationalizing the Sign of Entropy Change

Rationalizing the sign of the standard reaction entropy involves considering changes in the number of moles of gas and the phases of the compounds during a reaction. A negative entropy change typically indicates a decrease in disorder, such as when gas molecules combine to form fewer gas molecules or when a substance changes from a gaseous to a liquid state. Conversely, an increase in the number of gas molecules or a transition to a more disordered state results in a positive entropy change.

Transcript

00:01 In this question we've been asked to determine the standard entropy change of a reaction, standard entropy change.

00:07 And we know this to be the entropy change of the products, the standard entropy change of the products, minus that of the reactants, of the reactants.

00:20 And we always have to make sure that we include the stoichiometric coefficients, the stoichiometry coefficients of those products or reacting species in that reaction.

00:31 This is because the entropy, the standard entropy is given in joules per mole kelvin.

00:39 So we have to include this documentary coefficient to deal with this per mole so that at the end of the day we have a standard entropy change that is in joules per kelvin.

00:51 This value that we get from our appendices or from our tables is given per mole.

00:57 So whenever we have a more, for example, this documentary coefficient, being not equal to zero we have to multiply that stoichiometric by the value that we are retrieving from our tables for example say we have got a reacting with b to form c and d what we are going to be looking at we are going to be looking at the entropies of the products and subtract the entropy of the reactants this is what this formula is telling us now moving on to what we've been given the first reaction we're going to look at the products minus the reactants.

01:35 So the standard entropy change is going to be equal to one because we have one more of that substance, the product, multiplied by the entropy of that product being equal to 2, 29 .2 minus that of the reactants, which is to 1 .9 .3 minus 130 .7.

02:02 So the standard entropy change of that reaction is going to be equal to negative 2, negative 1 ,2008 .8, and this is in joel's per kelvin.

02:11 Now, if we are to look at this, this is telling us that the entropy change of this reaction is less than 0.

02:18 And this is happening because we have two moles of cassius products, cassius reactants, and we have one mole of the products.

02:30 So we've got two moles that are changing into one more.

02:34 So the entropy of this reaction is actually decreasing because the number of gaseous particles are decreasing.

02:41 As a result, the entropy change is going to be less than zero.

02:46 It makes sense to have an entropy change of negative 120 .8.

02:51 When we look at what is actually happening, the number of molecules are decreasing.

02:56 Remember, the entropy is a measure of the disorderness of a system.

03:00 And here we have two moles of gas, but finally we have one mole of gas.

03:05 So the disorderness of the system is going to decrease, and as a result, we are going to have a standard entropy change that is less than zero.

03:12 Hence this negative 1 -20 .8 jails per kelvin makes sense.

03:17 So moving on to the next one.

03:21 For the next system, we are going to say delta s standard being equal to 1, multiplied by 197 .7, plus one more...

Please give Ace some feedback