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Ch.5 - Thermochemistry
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All textbooksBrown 14th EditionCh.5 - ThermochemistryProblem 63

Chapter 5, Problem 63

Calculate the enthalpy change for the reaction P4O6(s) + 2 O2(g) → P4O10(s) given the following enthalpies of reaction: P4(s) + 3 O2(g) → P4O6(s) ΔH = -1640.1 kJ P4(s) + 5 O2(g) → P4O10(s) ΔH = -2940.1 kJ

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Hi everyone here we have a question that gives us the following infill piece of reactions. Software plus oxygen forms sulfur dioxide. And the entropy is negative. 296.8 killed joules per mole, sulfur plus three halves. Oxygen forms sulfur tri oxide and its entropy is negative 309 D 5.7 kg per mold. And our goal is to determine the entropy change for sulfur dioxide plus half oxygen forms sulfur trioxide. So let's see what we have here. We need each entropy to match our reaction. So for the first one we need to reverse it. So that would give us sulfur. Sulfur dioxide forms sulfur plus oxygen And its entropy will be 296 .8 killed Jules. Permal. For our 2nd 1 we need to just keep it the same So its entropy will stay negative, 395 .7 kg per mole. So now our entropy has gone to equal 296 0. kill jules. Permal plus negative .7 kg per mole And that equals 98 . kg Parimal. And that is our final answer. Thank you for watching. Bye.
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