At temperatures above 0°C, what is the sign of the change in free energy for the conversion of liquid water to ice?
A. Positive
B. Zero
C. Negative
D. Not known
dG = dHfus - Tfusion*dSfusion
0 = .....
Then dHfusion = Tfusion*dSfusion.
positive
To determine the sign of the change in free energy when liquid water converts to ice at temperatures above 0°C, we need to consider the conditions under which the conversion is occurring.
The change in free energy during a process can be calculated using the equation:
ΔG = ΔH - TΔS
Where:
ΔG is the change in free energy
ΔH is the change in enthalpy
T is the temperature in Kelvin
ΔS is the change in entropy
In the given scenario, liquid water is converting to ice at temperatures above 0°C. At temperatures above 0°C, the starting point is liquid water, and the final point is ice, which is a solid state.
The change in enthalpy (ΔH) for this process is negative since the conversion of liquid water to ice releases heat (exothermic process).
The change in entropy (ΔS) for this process is also negative as the water molecules become more ordered in the solid state.
Plugging the values into the equation, we have:
ΔG = ΔH - TΔS
Since both ΔH and ΔS are negative, the values of -TΔS and ΔH in the equation will determine the sign of ΔG.
If -TΔS is greater than ΔH, ΔG will be negative.
If -TΔS is less than ΔH, ΔG will be positive.
If -TΔS is equal to ΔH, ΔG will be zero.
Since both -TΔS and ΔH are negative in this case, the absolute value of -TΔS will be smaller than ΔH. Therefore, the change in free energy (ΔG) for the conversion of liquid water to ice at temperatures above 0°C will be positive.
So, the answer is A. Positive.