Given the following data, calculate ΔS°rxn (in J/K) for the following at 298 K:

3 A(s) + 2 B(g) → 2 C(g) ΔS°rxn = ???

S° (J/mol-K): A(s) = 63.6; B(g) = 196.3; C(g) = 246.4

Round your answer to ONE decimal place (i.e. 2.3) and do NOT include the units of J/K.
If your last digit is a zero after a decimal (i.e. 6.0), Carmen may drop the zero. That is OK!

10C

Question

Given the following data, calculate ΔS°rxn (in J/K) for the following at 298 K:

3 A(s) + 2 B(g) → 2 C(g)    ΔS°rxn = ???

S° (J/mol-K): A(s) = 63.6; B(g) = 196.3; C(g) = 246.4

Round your answer to ONE decimal place (i.e. 2.3) and do NOT include the units of J/K.
If your last digit is a zero after a decimal (i.e. 6.0), Carmen may drop the zero. That is OK!

10C

BlackTom AI · Accepted Answer

First, identify the reaction and list the standard molar entropies given: S°(A) = 63.6 J/mol·K, S°(B) = 196.3 J/mol·K, S°(C) = 246.4 J/mol·K. The reaction is 3 A(s) + 2 B(g) → 2 C(g). The standard entropy change for a reaction is ΔS°rxn = Σνp S°(products) − Σνr S°(reactants), where ν are the stoichiometric coefficients.
Next, calculate the total entropy of the products: 2 × S°(C) = 2 × 246.4 = 492.8 J/K.
Then calculate the total entropy of the reactants: (3 × S°(A)) + (2 × S°(B)) = 3 × 63.6 + 2 × 196.3 = 190.8 + 392.6 = 583.4 J/K.
Now subtract the reactants sum from the products sum to obtain ΔS°rxn: 492.8 − 583.4 = −90.6 J/K.
The problem asks for the value rounded to one decimal place and without units, so −90.6 becomes the final numeric result to one decimal place. The calculation shows the entropy decreases overall for this reaction, as indicated by the negative ΔS°rxn value.
If you were to verbally interpret: the production of fewer moles of gas on the product side and the involvement of a solid reactant contribute to the net decrease in disorder, consistent with the negative value.

CHEM 1220 SP2025 (18215) Exam 4- Requires Respondus LockDown Browser

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