To find the entropy change of the surroundings for the given process, we can use the following thermodynamic relation which connects the enthalpy change
(∆H) of a system to the entropy change of the surroundings (
∆Ssurroundings ) at constant temperature:
∆Ssurroundings =− In this formula,
∆H is the enthalpy change for the system, and
T is the temperature at which the process takes place. The negative sign indicates that if the process is endothermic
(∆H>0) for the system, the entropy of the surroundings decreases (
∆Ssurroundings <0 ), and vice versa for an exothermic process.
For the reaction given:
H2O(s)→H2O(ℓ) the enthalpy change is given as
+7kJmol−1 or
+7000Jmol−1 when converted to joules, since
1kJ=1000J, and the temperature is
300K. Plugging these values into our equation gives:
∆Ssurroundings =−To simplify this:
∆Ssurroundings =−JK−1mol−1∆Ssurroundings =−23.333...JK−1mol−1Therefore, the entropy change of the surroundings for the melting of ice at
300K with an enthalpy change of
+7kJmol−1 is approximately:
∆Ssurroundings ≈≈−23.3JK−1mol−1This corresponds to Option
B:−23.3JK−1.