8 P4010(s) + 6 H20(1) → 4 H3P04(1)Calculate the entropy change for thesystem. Show labelled work!b. Calculate the entropy change for thesurroundings (hint: find AH first). Showlabelled work!а.Calculate the total entropy change,ASuniverse, for this reaction (assumingС.T=25 °C)d. From a thermodynamics perspectivethen, based on your value forASuniverse are reactants more likely toproceed to products or remain asreactants?P4010(s)AH; (kJ mol1) = -2984.0S° (JK-1 mol-1) = 228.86

Since you have posted question with multiple subparts. So, as per guidelines i have solved first…

Answered: P4010(s) + 6 H20(1) → 4 H3P04(1)…

P4010(s) + 6 H20(1) → 4 H3P04(1) Calculate the entropy change for the system. Show labelled work! b. Calculate the entropy change for the surroundings (hint: find AH first). Show а.

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Crystal Structures

Solids are present all around us. They have different properties and applications. Diamond and graphite are two well-known solids. Diamond is the hardest material in the world and is used in making jewelry. Graphite on the other hand is soft. The lead of your pencil is graphite. Why are they so different in appearance and properties despite the fact they are chemically the same? Both are allotropes of carbon. The answer lies in the structure, the way constituent particles are arranged in the solids.

Question

P4010(s) + 6 H20(1) → 4 H3P04(1)
Calculate the entropy change for the
system. Show labelled work!
b. Calculate the entropy change for the
surroundings (hint: find AH first). Show
labelled work!
а.
Calculate the total entropy change,
ASuniverse, for this reaction (assuming
С.
T=25 °C)
d. From a thermodynamics perspective
then, based on your value for
ASuniverse are reactants more likely to
proceed to products or remain as
reactants?
P4010(s)
AH; (kJ mol1) = -2984.0
S° (JK-1 mol-1) = 228.86

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