How would you calculate the standard entropy change (J/mol-K) here?

For the reaction whose reactants and products are:











reactants: C (graphite) CO2 (g)








products: CO (g)








calculate the standard entropy change (J/mol-K).





The answer depends on how you balance the equation. For this question balance the equation using the smallest ratio of WHOLE numbers.





Enter a numeric answer only








ANy help would be greatly appreciated!!!

How would you calculate the standard entropy change (J/mol-K) here?
Balanced reaction would be:





C (s, graphite) + CO2 (g) -%26gt; 2 CO (g)





If you use the standard entropies provided in the appendix of a textbook:





delta S,r = products - reactants


= 2 ( 197.67 J / mol K) - [ 1 (213.74 J / mol K) + 1 ( 5.740 J / mol K)


= 395.34 J / mol K - 219.48 J / mol K


= 175.86 J / mol K





That is, for every mole of CO2 or C reacted, the entropy of the system increases by 175.86 J / mol K. Dunno if you want per mole product, or whatever.





This may vary slightly depending on the values provided for Sm in your textbook.
Reply:Well, first the balanced equation:


2 C + CO2 = 2 CO





Equation: Q (heat) = (m)(c)(DeltaT)





Solving for c (entropy change in J/grams/K):


c = (mass)(Delta T)/Q