A) \[\Delta H=0\]and\[\Delta S<0\]
B) \[\Delta H>0\]and \[\Delta S<0\]
C) \[\Delta H>0\]and\[\Delta S<0\]
D) \[\Delta H>0\]and\[\Delta S<0\]
Correct Answer: B
Solution :
Key Idea: \[\Delta H=\Delta E+\Delta nRT\] \[\Delta n=\] number of moles of product - number of moles of reactants\[\Delta G=\Delta H-T\Delta S\] For a spontaneous process AC must be negative. \[PC{{l}_{5}}(g)PC{{l}_{3}}(g)+C{{l}_{2}}(g)\]In this reaction\[\Delta n=2-1=1\] Thus, \[\Delta H\] is positive, ie, > 0 If \[\Delta H\] is positive, then to maintain the value of \[\Delta G\] negative, \[\Delta S\] should be positive, ie, \[\Delta S>0.\]You need to login to perform this action.
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