Find ΔG and ΔA for the freezing of 1.00 mol of H20 at 0°c and 1atm. At 0 ° C and 1 atm the density of ice and liquid water are 0.917 g / mL and 1 g / mL, respectively.
Well, if you're looking for ΔG and ΔA for the freezing of 1.00 mol of H2O at 0°C and 1 atm, you're in for a chilly treat!
First, let's start with ΔG, which represents the change in Gibbs free energy. The equation for ΔG is ΔG = ΔH - TΔS, where ΔH is the change in enthalpy, T is the temperature in Kelvin, and ΔS is the change in entropy.
For the freezing of H2O, ΔH is the enthalpy of fusion, which is the amount of heat released when a substance changes from a liquid to a solid at a constant temperature. In this case, it's 0°C. Since we know that the density of ice is 0.917 g/mL and the molar mass of water is 18.02 g/mol, we can calculate the moles of water in 1 mL of ice: 1 mL * (1 g / 0.917 g/mL) / 18.02 g/mol = 0.055 mol.
So, ΔH = -(-6.01 kJ/mol) = 6.01 kJ/mol (where -6.01 kJ/mol is the negative value of the enthalpy of fusion of water).
Next, we'll calculate ΔS, which is the change in entropy. For this calculation, we need to know the change in volume between liquid water and ice. Since the density of liquid water is 1 g/mL and the density of ice is 0.917 g/mL, we can say that the decrease in volume during freezing is 1 mL - 0.917 mL = 0.083 mL.
Since 1 mol of water occupies 18.02 mL, the decrease in volume for 1 mol of water during freezing is 0.083 mL/mol * (1 mol / 18.02 mL) = 0.0046 mol.
Now, we have all the values we need to calculate ΔG. Let's plug them into the equation: ΔG = ΔH - TΔS.
Assuming a temperature of 0°C = 273 K, ΔG = 6.01 kJ/mol - 273 K * (0.0046 mol/K) = 4.77 kJ/mol.
Now, onto ΔA, which is the change in Helmholtz free energy. ΔA is given by ΔA = ΔU - TΔS, where ΔU is the change in internal energy.
For the freezing of H2O, ΔU is the change in internal energy, which is also equal to the enthalpy change, since there is no expansion or compression in this process. So, ΔU = ΔH = 6.01 kJ/mol.
Plugging in the values, ΔA = 6.01 kJ/mol - 273 K * (0.0046 mol/K) = 4.77 kJ/mol.
So, ΔG = ΔA = 4.77 kJ/mol. Now that's a cool result!
I hope I didn't leave you feeling too frosty with all those calculations.
To find ΔG (change in Gibbs free energy) and ΔA (change in Helmholtz free energy) for the freezing of 1.00 mol of H2O at 0°C and 1 atm, we'll need to use the following formulas:
ΔG = ΔH - TΔS
ΔA = ΔU - TΔS
where:
ΔH = enthalpy change
ΔU = internal energy change
T = temperature in Kelvin
ΔS = entropy change
First, let's find the enthalpy change (ΔH) for the freezing of 1.00 mol of H2O. The enthalpy change can be obtained from the heat of fusion of water, which is 6.01 kJ/mol.
Next, we need to find the entropy change (ΔS) for the freezing process. The entropy change can be calculated using the formula:
ΔS = (ΔH / T) + ΔS(sys)
where ΔS(sys) is the change in entropy of the system.
For the freezing of H2O, ΔS(sys) can be determined using the formula:
ΔS(sys) = ∆S(fus) = ΔH / T
Where ∆S(fus) is the entropy change associated with the fusion of water, which is 0.0228 kJ/(mol⋅K).
Now, let's find ΔG and ΔA using the formulas mentioned earlier.
ΔG = ΔH - TΔS
ΔA = ΔU - TΔS
Since for this process, ΔU is approximately zero, we can write:
ΔG = ΔH - TΔS
ΔA = -TΔS
Substituting the values we know:
ΔH = 6.01 kJ/mol
T = 0°C + 273.15 = 273.15 K
ΔS = ∆S(fus) = 0.0228 kJ/(mol⋅K)
ΔG = 6.01 kJ/mol - 273.15 K * 0.0228 kJ/(mol⋅K)
ΔA = -273.15 K * 0.0228 kJ/(mol⋅K)
Calculating ΔG and ΔA:
ΔG = 6.01 kJ/mol - 6.232 kJ/mol
= -0.222 kJ/mol
ΔA = -273.15 K * 0.0228 kJ/(mol⋅K)
= -6.232 kJ/mol
Therefore, ΔG for the freezing of 1.00 mol of H2O at 0°C and 1 atm is approximately -0.222 kJ/mol, and ΔA is approximately -6.232 kJ/mol.
To find ΔG and ΔA for the freezing of 1.00 mol of H2O at 0°C and 1 atm, we need to use the equation:
ΔG = ΔH - TΔS
where ΔH is the enthalpy change, T is the temperature in Kelvin, and ΔS is the entropy change.
Step 1: Determine the enthalpy change (ΔH):
The enthalpy change for the freezing of water is the heat released when 1.00 mol of water freezes. The enthalpy of fusion (ΔH) of water is 6.01 kJ/mol.
Step 2: Convert the temperature to Kelvin (K):
0°C is equivalent to 273.15 K.
Step 3: Determine the entropy change (ΔS):
The entropy change (ΔS) can be found using the equation:
ΔS = ΔSfus(T) + ΔSvap(T)
where ΔSfus(T) is the entropy change during fusion and ΔSvap(T) is the entropy change during vaporization. Since we are dealing with the freezing of water, we only need to consider ΔSfus(T) for this calculation.
The entropy change during fusion of water is given by the equation:
ΔSfus(T) = ΔHfus / T
where ΔHfus is the enthalpy of fusion and T is the temperature in Kelvin.
Step 4: Calculate ΔS:
Substituting the known values, we have:
ΔS = (ΔHfus) / T = 6.01 kJ/mol / 273.15 K
Step 5: Calculate ΔG:
Substitute the values of ΔH, T, and ΔS into the equation:
ΔG = ΔH - TΔS
and calculate ΔG.
Step 6: Calculate ΔA:
Since ΔA is the Helmholtz free energy, it can be calculated using the equation:
ΔA = ΔU - TΔS
where ΔU is the internal energy change.
Alternatively, ΔA can also be calculated using the equation:
ΔA = ΔG + PΔV
where P is the pressure and ΔV is the change in volume.
However, in this case, since the pressure is constant at 1 atm and there is no change in volume during freezing, we can simplify the equation to:
ΔA = ΔG
Therefore, ΔA can be calculated using the value of ΔG obtained in the previous step.
Note:
You can find the value of the enthalpy change (ΔH) and entropy change (ΔS) from thermodynamic databases or reference books. These values are experimentally determined and can vary depending on the source.