Predict the sign of ΔS^∘ and then calculate ΔS^∘ for each of the following reactions. a. H2(g)+

Predict the sign of ΔS^∘ and then calculate ΔS^∘ for each of...

Key Concepts

Standard Molar Entropy

Standard molar entropy is a thermodynamic property that quantifies the absolute entropy of a substance under standard conditions (usually 1 atm and 298 K). It is a measure of the disorder or randomness of a system and is typically expressed in units of joules per mole per kelvin (J/(mol·K)). These values are tabulated for various substances and serve as the basis for calculating changes in entropy for chemical reactions.

Entropy Change (?S°)

The standard entropy change of a reaction (?S°) is determined by taking the sum of the standard molar entropies of the products (multiplied by their stoichiometric coefficients) and subtracting the sum of the standard molar entropies of the reactants. This calculation follows the formula ?S° = ?(n·S°_products) - ?(n·S°_reactants), providing insights into the overall increase or decrease in randomness during a reaction.

Phase Effects on Entropy

The physical state of substances significantly affects their entropy. Generally, gases possess much higher entropy than liquids and solids because gas molecules have more freedom of movement, resulting in a higher degree of disorder. In reactions, a decrease in the number of gaseous molecules or a conversion of gas to liquid usually implies a decrease in entropy, whereas an increase in gas molecules or transition from solid to gas indicates an entropy increase.

Reaction Stoichiometry

Reaction stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It plays an essential role in entropy calculations because each term in the entropy change formula is multiplied by its respective stoichiometric coefficient. Accurate stoichiometric balancing ensures that the contribution of each substance's standard entropy is correctly accounted for when determining the overall ?S° of the reaction.

Ionic Solvation and Entropy

When a compound dissolves in water and dissociates into ions, hydration (or solvation) processes occur, which can reduce the system's entropy. The ordering of water molecules around the ions, due to electrostatic interactions, generally leads to a decrease in randomness compared to the undissociated gas phase. This concept is critical for understanding entropy changes in reactions that involve the formation of aqueous ions.

Transcript

00:01 Hi guys.

00:03 So in problem 50 we are asked to predict the sign and calculate standard entropys for the following reactions.

00:22 So let's start with the first one which is a to predict the entropy for the first one we look at the reactants the product so moving from the reactant to the product we're moving from gaseous face to liquid face and what that does is to decrease the disorderliness of the system and by so doing we're decreasing the entropy so the entropy is negative so let's calculate that two calculate that we know that entropy is a state function which means it depends only on the initial and final states of the system so we have the sum of the entropies of the product minus the sum of the intropies of the reactants so we can look up for the entropies of each of the involving species from the standard tables pretty much in any at the back of any physical chemistry textbook these are standards and they are tabulated and we can find them at appendix so for the product in this case we have just one so which is 70 um forgive me so we need to include the most so how many more of the product we have is one right so we have one more of that which is more of water right and it has 70 you per kelvin more minus one more of hydrogen right got this is liquid we should should indicate the states as well this is gas which has to go five jou per kelvin mole so this is plus more of oxygen and oxygen has 131 jou per kelvin more so in each case the mole will cancel out in each case the mole will cancel out right and then when you do this you will get negative 164 jr per kelvin because the more is gone so you can see that we predicted delta s to be negative and of course it's negative so let's look at this second problem the second problem has gases in both reactants and product side so everything in the system is gaseous both reactions and product so in that case you some of the most of the reactants and then some of the most of the product and compare so here we have two plus three which is five moles of gaseous molecules and here we have 2 plus 4, 6 moles of gaseous molecules.

05:47 So because we have more moles, the gaseous molecules in the product side, it means that there's an increase in this orderliness of the system, and hence delta s is positive, because it's increased.

06:17 So that is the prediction.

06:20 And let's do the same and see what we'll get.

06:25 Remember, we said delta s is given by the summation of the entropy of the product minus summation of the entropy of the reactant.

06:37 So the entropy of the product, starting with carbon for oxide, have two moles, co2 gas, and then co2 has two.

06:55 One full remember the unit is joe pa kelvin mole for all of them so i may not show them all but the unit is the same so plus this is for um four malls of water right gas multiplied by one eight nine joe pi kelvin mole right minus then that of reactant we have two moles methanol gas and that will be two o five right zero per kelvin more plus three most oxygen gas um so i think the the death of methanol is 240 in this case is 240 and that of oxygen is 204 205 so 240 and then that of oxygen is 205 05 jou per kelvin mole so again the mole councils out more councils out more council's out so when you compute for the delta s...

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