Consider the freezing of liquid water at -10*C. For this process what are the si
ID: 827586 • Letter: C
Question
Consider the freezing of liquid water at -10*C. For this process what are the signs for delta h, delta S and delta G?
delta H delta S delta G
a. + - 0
b. - + 0
c. - + -
d. + - -
e. - - -
I think it is e. Is this right. if not what is the right answer.
Explanation / Answer
yes E is the correct answer
reason : 1] for delta H: freezing is an exothermic process , as it involves formation of more stable bonds in water , hydrogen bond formation,whenever more stable bonds are formed energy is released, so -ve change in enthalpy
2] entropy of solid are always less than entropy of liquid so freezing always result in -ve change in entropy ,enetropy is a measure of randomness in motion of particles , higher the randomness , higher is entropy , when in solid state particles cant more so randomly while in liquid phase particles can flow easily , move easily , so there entropy higher
3]using delta G = delta H - T[delta S ]
experimental values of entropy change and enthalpy change shows its a spontaneous process so delta G -ve