Given the options, and entropy data, which gives the correct #DeltaS^@_"rxn"# for the reaction.... #2NH_3(g)rarrN_2(g)+3H_2(g)# #"A. -299"JK^-1mol^-1# #"B. -199"JK^-1mol^-1# #"C. +199"JK^-1mol^-1# #"D. +99"JK^-1mol^-1#?

Answer 1

This is simple summation.....I gets #C#

Look at your reaction:

#2NH_3(g) rarr N_2(g) + 3H_2(g)#

You produce 4 moles of gas from 2 moles of gas; the entropy, the statistical probability for disorder is likely to INCREASE......and so that leaves only #B# and #C# as viable answers.....

To decide, we have to an entropy summation......

#DeltaS_"rxn"^@=Sigma{S_"products"^@-S_"reactants"^@}#

But we have been given the individual entropies....and so.....

#DeltaS_"rxn"^@={3xx131+192-2xx193}*J*K^-1*mol^-1#

#=199*J*K^-1*mol^-1#, where #mol^-1# means per mole of REACTION as written.........

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Answer 2

Luna Chen

The correct answer is: C. +199 JK^-1mol^-1

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Given the options, and entropy data, which gives the correct #DeltaS^@_"rxn"# for the reaction.... #2NH_3(g)rarrN_2(g)+3H_2(g)# #"A. -299"*J*K^-1*mol^-1# #"B. -199"*J*K^-1*mol^-1# #"C. +199"*J*K^-1*mol^-1# #"D. +99"*J*K^-1*mol^-1#?

Given the options, and entropy data, which gives the correct #DeltaS^@_"rxn"# for the reaction.... #2NH_3(g)rarrN_2(g)+3H_2(g)# #"A. -299"JK^-1mol^-1# #"B. -199"JK^-1mol^-1# #"C. +199"JK^-1mol^-1# #"D. +99"JK^-1mol^-1#?