xyM Homework #7 M previous Attempt View x nof A Strong Acid x fezto.mheducation.

Question

xyM Homework #7 M previous Attempt View x nof A Strong Acid x fezto.mheducation.com/hm.tpx . à ezto.mhed uoregon.edu Blackboard Webmail Q Webmail M CH221, 222, 223 Onl...Canvas uoregon.edu Blackboard Mend E connect General Chemistr CRN 31193 HEMISTRY work #7 ·1 Question #7 (of 15) value 10.00 points Question Be sure to answer all parts. FindS" for the combustion of ammonia to nitrogen dioxide and water vapor. J/K Is this the sign of AS you would get if you only considered the change in moles of gas during the reaction? yes no

Explanation / Answer

The reaction of Ammonia undergoing combustion to give NO2 and H2O is

4NH3+7O2----->4NO2+6H2O

From literature

Entropy of NH3= 192.45 J/K mol

Entropy of O2 = 205 j/K mol

Entropy of NO2=240 J/K mol

Entropy of H2O=189 j/K mol

Since entropy is expressed in terms of moles, we have to account for stoichiometric coefficients

Standard Entropy change of reaction= Entropy of NO2*its stoichiometric coefficient+ Entropy of H2O * its stoichiometric coefficient- (Entropy of NH3* its stoichiomteric coefficient+Entropy of Oxygen* its stoichiometric coefiicient

Standard entropy change of reaction =6*189+4*240-(4*192.45+7*205)=-110.8 j/Kmol

Entropy change is -ve

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