In the previous reaction of carbon with water, why does the standard entropy you calculated have the arithmetic sign that it does (zero/positive/negative)?C(s) + H2O(g) <--> CO(g) + H2(g)SubstanceS° (J/mol*K)C(s)4.7H2O(g)14CO(g)57H2(g)119 A.That value is negative in value because there are more gaseous product molecules than there are gaseous reactant molecules. B.That value is positive because C(s) is an exceptional case to the generalized Second Law of Thermodynamics. C.That value is negative in value because the temperature of the reaction is below the critical point for gaseous carbon monoxide. D.That value is positive in value because there are more gaseous product molecules than there are gaseous reactant molecules

The balanced given reaction is as follows,

Answered: In the previous reaction of carbon with…

Chemistry: An Atoms First Approach

2nd Edition

ISBN:9781305079243

Author:Steven S. Zumdahl, Susan A. Zumdahl

Publisher:Steven S. Zumdahl, Susan A. Zumdahl

Chapter16: Spontaneity, Entropy, And Free Energy

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Problem 84AE: Calculate the entropy change for the vaporization of liquid methane and liquid hexane using the...

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In the previous reaction of carbon with water, why does the standard entropy you calculated have the arithmetic sign that it does (zero/positive/negative)?

C(s) + H₂O(g) <--> CO(g) + H₂(g)

Substance	S° (J/mol*K)
C(s)	4.7
H₂O(g)	14
CO(g)	57
H₂(g)	119

	A.	That value is negative in value because there are more gaseous product molecules than there are gaseous reactant molecules.
	B.	That value is positive because C(s) is an exceptional case to the generalized Second Law of Thermodynamics.
	C.	That value is negative in value because the temperature of the reaction is below the critical point for gaseous carbon monoxide.
	D.	That value is positive in value because there are more gaseous product molecules than there are gaseous reactant molecules

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