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Ch.5 - Thermochemistry
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All textbooksBrown 14th EditionCh.5 - ThermochemistryProblem 66

Chapter 5, Problem 66

Given the data N2(g) + O2(g) → 2 NO(g) ΔH = +180.7 kJ 2 NO(g) + O2(g) → 2 NO2(g) ΔH = -113.1 kJ 2 N2O(g) → 2 N2(g) + O2(g) ΔH = -163.2 kJ use Hess's law to calculate ΔH for the reaction N2O(g) + NO2(g) → 3 NO(g)

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Hi everyone here we have a question asking us to determine the entropy of the reaction. Using Hess's law. Iron sulfide plus five halves, oxygen forms iron oxide plus two sulfur oxide. Given the following data. So let's go through each of these. So we have iron plus two sulfur forms iron sulfate and its entropy is negative 178.2 kg joules per mole. So we need to reverse that to match our reaction. So that is 178 point to kill jules Permal. And then we have iron plus half oxygen forms iron oxide and its entropy is negative 272.0 kg joules per mole. And we're just going to keep that one the same. So that will be negative, 272 zero kg Permal. And lastly we have sulfur plus oxygen forms sulfur dioxide And it's info p is negative, 296.8 kg joules per mole. And we need to multiply by that by two to match our reaction. So that will be negative .6 kg per mole. So now we have to add this all up to get our entropy. So that's going to equal 178 points to kill jewels per mole plus - zero killed jules Permal plus -593 . Killed Jules, Caramel. And that equals negative .4 Kill Jules Permal. And that is our final answer. Thank you for watching. Bye
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Textbook Question

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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Textbook Question

From the enthalpies of reaction 2 C(s) + O2(g) → 2 CO(g) ΔH = -221.0 kJ 2 C(s) + O2(g) + 4 H2(g) → 2 CH3OH(g) ΔH = -402.4 kJ Calculate ΔH for the reaction CO(g) + 2 H2(g) → CH3OH(g)

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Textbook Question

From the enthalpies of reaction H2(g) + F2(g) → 2 HF(g) ΔH = -537 kJ C(s) + 2 F2(g) → CF4(g) ΔH = -680 kJ 2 C(s) + 2 H2(g) → C2H4(g) ΔH = +52.3 kJ Calculate H for the reaction of ethylene with F2: C2H4(g) + 6 F2(g) → 2 CF4(g) + 4 HF(g)

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