Example 16.4: Will Ice Spontaneously Melt?
The entropy change for the process
\(\ce{H2O(s)}\)\(\ce{->}\)\(\ce{H2O(l)}\)\(\ce{ }\)
is
22.1 J/K and requires that the surroundings transfer
6.00 kJ of heat to the system. Is the process spontaneous at
-10.00 °C? Is it spontaneous at
+10.00 °C?
Solution
\(ΔS_{\mathrm{sys}}\) \(= 22.1\ \frac{\mathrm{J}}{\mathrm{K}}\)
\(q_{\mathrm{surr}}\) \(= -6.00\ \mathrm{kJ}\)
\(\mathrm{ΔSsurr} = \dfrac{q_{\mathrm{surr}}}{T}\)
\(T_{\mathrm{2}}\) \(= 10.00\ \mathrm{°aC}\)
We can assess the spontaneity of the process by calculating the entropy change of the universe. If
\(\mathrm{ΔSuniv}\) is positive, then the process is spontaneous. At both temperatures,
\(\mathrm{ΔSsys}\) \( = 22.1\ \frac{\mathrm{J}}{\mathrm{K}}\) and qsurr
\( = - 6.00\) kJ.
At
-10.00 °C (
263.15 K), the following is true:
\(ΔS_{\mathrm{surr\ 1}}\) \(= \dfrac{q_{\mathrm{surr}}}{T_{\mathrm{1}}}\)
\(\ \ \ =\dfrac{-6.00\ \mathrm{kJ}}{263.15\ \mathrm{K}}\)
\(\ \ \ =-0.02280\ \frac{\mathrm{kJ}}{\mathrm{K}}\)
\(ΔS_{\mathrm{univ}}\) \(= ΔS_{\mathrm{sys}} + ΔS_{\mathrm{surr\ 1}}\)
\(\ \ \ =22.1\ \frac{\mathrm{J}}{\mathrm{K}} + -0.02280\ \frac{\mathrm{kJ}}{\mathrm{K}}\)
\(\ \ \ =-0.7\ \frac{\mathrm{J}}{\mathrm{K}}\)
\(ΔS_{\mathrm{univ}}\) <
0, so melting is nonspontaneous (not spontaneous) at
-10.0 °C.
At
10.00 °C (
283.15 K), the following is true:
\(ΔS_{\mathrm{surr\ 2}}\) \(= \dfrac{q_{\mathrm{surr}}}{T_{\mathrm{2}}}\)
\(\ \ \ =\dfrac{-6.00\ \mathrm{kJ}}{283.15\ \mathrm{K}}\)
\(\ \ \ =-0.02119\ \frac{\mathrm{kJ}}{\mathrm{K}}\)
\(ΔS_{\mathrm{univ\ 2}}\) \(= ΔS_{\mathrm{sys}} + ΔS_{\mathrm{surr\ 2}}\)
\(\ \ \ =22.1\ \frac{\mathrm{J}}{\mathrm{K}} + -0.02119\ \frac{\mathrm{kJ}}{\mathrm{K}}\)
\(\ \ \ =0.9\ \frac{\mathrm{J}}{\mathrm{K}}\)
\(\mathrm{ΔSuniv}\) >
0, so melting is spontaneous at
10.00 °C.