Given that ΔH (fusion) of water is 6025 J/mol, Cp,m(ice) = 37.7 J K–1 mol–1, and Cp,m(water,liquid) = 75.3 J K–1 mol–1, calculate ΔG (in J) for the freezing of 89 moles of water at 1 atm pressure and the constant temperature T = 14.4 °C.
I got -4458.92J, but it's wrong. Can anyone show me how to approach this question? Thank you!
ΔG = ΔH - TΔS
ΔH = 6025 J/mol
T = 14.4 °C = 287.6 K
ΔS = Cp,m(ice) - Cp,m(water,liquid)
= 37.7 J K–1 mol–1 - 75.3 J K–1 mol–1
= -37.6 J K–1 mol–1
ΔG = 6025 J/mol - (287.6 K)(-37.6 J K–1 mol–1)
= 6025 J/mol + 10862.56 J/mol
= 16887.56 J/mol
ΔG = (16887.56 J/mol)(89 mol)
= 1508982.04 J
To calculate ΔG for the freezing of water, we will use the equation:
ΔG = ΔH - TΔS
Where ΔH is the enthalpy change, T is the temperature in Kelvin, and ΔS is the entropy change.
First, we need to calculate the entropy change using the equation:
ΔS = ΔS(liquid) - ΔS(ice)
ΔS(liquid) = Cp,m(liquid) * ln(T2/T1)
ΔS(ice) = Cp,m(ice) * ln(T2/T1)
T1 = 14.4 °C = 14.4 + 273.15 = 287.55 K
T2 = 0 °C = 0 + 273.15 = 273.15 K
ΔS(liquid) = 75.3 J K^(-1) mol^(-1) * ln(273.15/287.55)
ΔS(ice) = 37.7 J K^(-1) mol^(-1) * ln(273.15/287.55)
Now we can calculate ΔS:
ΔS = ΔS(liquid) - ΔS(ice)
Next, we need to convert the temperature from Celsius to Kelvin:
T = 14.4 °C = 14.4 + 273.15 = 287.55 K
Now we can calculate ΔG:
ΔG = ΔH - TΔS
Finally, to find ΔG for the freezing of 89 moles of water at 1 atm pressure, we multiply ΔG by the number of moles:
ΔG (total) = ΔG * moles
Note: Make sure to use the correct units for the calculations.
Using these steps, let's calculate ΔG for the freezing of water.
To approach this question, we need to use the Gibbs Free Energy equation:
ΔG = ΔH - TΔS
Where:
ΔG is the change in Gibbs Free Energy
ΔH is the enthalpy change
T is the temperature in kelvin
ΔS is the change in entropy
First, let's calculate the entropy change (ΔS).
ΔS = Cp,m(liquid) * ln(T2/T1) - R * ln(P2/P1)
Where:
Cp,m(liquid) is the molar heat capacity of the liquid water
ln is the natural logarithm
T2/T1 is the ratio of final and initial temperatures
R is the ideal gas constant (8.314 J K^(-1) mol^(-1))
ln(P2/P1) is the ratio of final and initial pressures
Let's calculate ΔS.
Given:
Cp,m(liquid) = 75.3 J K^(-1) mol^(-1)
T2 = 14.4 °C + 273.15 = 287.55 K (temperature in kelvin)
T1 = 0 °C + 273.15 = 273.15 K (initial temperature)
P2/P1 = 1 (since the pressure remains constant)
ΔS = 75.3 * ln(287.55/273.15) - 8.314 * ln(1)
= 75.3 * ln(1.052) - 8.314 * ln(1)
= 75.3 * 0.0516 - 8.314 * 0
= 3.874 - 0
= 3.874 J K^(-1) mol^(-1)
Next, let's calculate ΔG using the ΔG equation:
ΔG = ΔH - TΔS
Given:
ΔH = 6025 J/mol (enthalpy change)
T = 14.4 °C + 273.15 = 287.55 K (temperature in kelvin)
ΔS = 3.874 J K^(-1) mol^(-1) (entropy change)
ΔG = 6025 - 287.55 * 3.874
= 6025 - 1113.8397
= 4907.1603 J
Therefore, the ΔG value for the freezing of 89 moles of water at 1 atm pressure and T = 14.4 °C is approximately 4907.16 J.