Question

The melting point of water is 0°C at 1 atm pressure because under these conditions:

A. ΔS for the process H2O(s) → H2O(l) is positive.
B. ΔS and ΔSsurr for the process H2O(s) → H2O(l) are both positive.
C. ΔS and ΔSsurr for the process H2O(s) → H2O(l) are equal in magnitude and opposite in sign.
D. ΔG is positive for the process H2O(s) → H2O(l).
E. None of these is correct.

Answer 1

The correct answer is option C, that is, ΔS and ΔSsurr for the process H2O (s) ⇒ H2O(l) are equal in magnitude and opposite in sign.

Step-by-step explanation:

The temperature at which solid state of water get transformed into liquid state is termed as the melting point of 0 °C. It can be shown by the reaction:

H2O (s) ⇒ H2O (l)

The degree of randomness of a molecule is known as entropy. With the transformation of ice into liquid state, there is an increase in randomness. Thus, the value of entropy becomes positive as shown:

Entropy change (ΔSsys) = ΔSproduct - ΔSreactant

= (69.9 - 47.89) J mol/K

= 22.0 J mol/K

Therefore, the value of entropy change is positive.

Now the value of entropy for surrounding ΔSsurr will be,

ΔSsurr = -ΔHfusion/T

= -6012 j/mol/273

= -22.0 J/molK

Hence, the value of ΔSsurr and ΔSsys exhibit same magnitude with opposite sign.