a) The vapour pressure of mercury at 100 ºC is 36.38 Pa. Calculate ∆Gº of vaporization of Hg (ℓ) at 100 º C.
b) The vapour pressure of acetone, CH3COCH3 (ℓ) at 25 º C is 232 Torr. If ∆Gf º of CH3COCH3 (ℓ) at 25ºC is −155.7 kJ/mol , calculate ∆Gf º of CH3COCH3 (g) at 25 ºC.
c) For the reaction 2 CuBr2 (s) → 2 CuBr (s) + Br2 (g) , the equilibrium vapour pressure of Br2 (g) is 1.90 x 10− 6 kPa at 298 K. Calculate ∆G for this reaction at 298 K if Br2 (g) is produced at a pressure of 1.00 x 10− 7 kPa.
d) The equilibrium constant for the reaction HNO2 (aq) + H2 O ( ℓ) ↔ NO2¯ (g) + H3O+ (aq)
is 4.3 x 10 − 4 at 25º C. Will nitrous acid spontaneously dissociate when:
at [HNO2] = [NO2¯] = [H3O + ] = 1.0 M ?
at [HNO2] = 1.0 M and [NO2¯ ] = [H3O + ] = 1.0 x 10− 5 M ?
thanks in advance!
a) The formula to calculate the standard Gibbs free energy (∆Gº) of vaporization is:
∆Gº = -RT ln(P)
Where:
∆Gº is the standard Gibbs free energy in J/mol
R is the ideal gas constant (8.314 J/(mol·K))
T is the temperature in Kelvin
P is the vapor pressure in Pa
Given that the vapor pressure of mercury at 100 ºC is 36.38 Pa, we can calculate ∆Gº as follows:
∆Gº = - (8.314 J/(mol·K)) * (373 K) * ln(36.38 Pa)
∆Gº ≈ - (8.314 J/(mol·K)) * (373 K) * (-3.308)
∆Gº ≈ 9,800 J/mol
Therefore, the ∆Gº of vaporization of Hg (ℓ) at 100 ºC is approximately 9,800 J/mol.
b) To calculate the ∆Gf º of CH3COCH3 (g) at 25 ºC, we can use the equation:
∆Gf º (g) = ∆Gf º (ℓ) + RT ln(P)
Where:
∆Gf º (g) is the standard Gibbs free energy of formation of CH3COCH3 (g)
∆Gf º (ℓ) is the standard Gibbs free energy of formation of CH3COCH3 (ℓ)
R is the ideal gas constant (8.314 J/(mol·K))
T is the temperature in Kelvin
P is the vapor pressure in Pa
Given that ∆Gf º (ℓ) of CH3COCH3 at 25 ºC is -155.7 kJ/mol and the vapor pressure of CH3COCH3 (ℓ) at 25 ºC is 232 Torr (which is equivalent to 30702.4 Pa), we can calculate ∆Gf º (g) as follows:
∆Gf º (g) = -155.7 kJ/mol + (8.314 J/(mol·K)) * (298 K) * ln(30702.4 Pa)
∆Gf º (g) ≈ -155,700 J/mol + (8.314 J/(mol·K)) * (298 K) * (10.332)
∆Gf º (g) ≈ - 155,700 J/mol + 24,699 J/mol
∆Gf º (g) ≈ - 131,001 J/mol
Therefore, the ∆Gf º of CH3COCH3 (g) at 25 ºC is approximately -131,001 J/mol.
c) To calculate ∆G for the reaction 2 CuBr2 (s) → 2 CuBr (s) + Br2 (g), we need to use the relationship between the Gibbs free energy and the equilibrium constant:
∆G = -RT ln(K)
Where:
∆G is the Gibbs free energy in J/mol
R is the ideal gas constant (8.314 J/(mol·K))
T is the temperature in Kelvin
K is the equilibrium constant
Given that the equilibrium vapor pressure of Br2 (g) is 1.90 x 10^−6 kPa at 298 K (which is equivalent to 1.90 x 10^−11 Pa), and the reaction produces Br2 (g) at a pressure of 1.00 x 10^−7 kPa (which is equivalent to 1.00 x 10^−12 Pa), we can calculate ∆G as follows:
∆G = - (8.314 J/(mol·K)) * (298 K) * ln((1.90 x 10^−11) / (1.00 x 10^−12))
∆G ≈ - (8.314 J/(mol·K)) * (298 K) * (-2.988)
∆G ≈ 7,650 J/mol
Therefore, ∆G for the reaction at 298 K is approximately 7,650 J/mol.
d) To determine if nitrous acid will spontaneously dissociate under certain conditions, we need to compare the calculated ∆G value to zero (∆G < 0 for spontaneous process).
at [HNO2] = [NO2¯] = [H3O + ] = 1.0 M:
To calculate ∆G for this condition, we can use the equation:
∆G = -RT ln(K)
Where:
∆G is the Gibbs free energy in J/mol
R is the ideal gas constant (8.314 J/(mol·K))
T is the temperature in Kelvin
K is the equilibrium constant
Given that the equilibrium constant for the reaction HNO2 (aq) + H2 O ( ℓ) ↔ NO2¯ (g) + H3O+ (aq) is 4.3 x 10^−4 at 25º C, we can calculate ∆G as follows:
∆G = - (8.314 J/(mol·K)) * (298 K) * ln(4.3 x 10^−4)
∆G ≈ - (8.314 J/(mol·K)) * (298 K) * (-7.758)
∆G ≈ 18,760 J/mol
Since ∆G is greater than zero, nitrous acid will not spontaneously dissociate under the given conditions.
at [HNO2] = 1.0 M and [NO2¯] = [H3O + ] = 1.0 x 10^−5 M:
Using the same equation as before, we can calculate ∆G for this condition:
∆G = - (8.314 J/(mol·K)) * (298 K) * ln(4.3 x 10^−4)
∆G ≈ - (8.314 J/(mol·K)) * (298 K) * (-13.270)
∆G ≈ 34,101 J/mol
Again, since ∆G is greater than zero, nitrous acid will not spontaneously dissociate under these conditions either.
a) To calculate ∆Gº of vaporization of Hg (ℓ) at 100 ºC, you can use the equation ∆Gº = -RT ln(P), where R is the gas constant (8.314 J/mol·K), T is the temperature in Kelvin, and P is the vapor pressure in Pascal.
First, convert the temperature from Celsius to Kelvin by adding 273.15:
T = 100 ºC + 273.15 = 373.15 K
Then, plug in the values into the equation:
∆Gº = - (8.314 J/mol·K) * 373.15 K * ln(36.38 Pa)
This will give you the value of ∆Gº in J/mol.
b) To calculate ∆Gf º of CH3COCH3 (g) at 25 ºC, you can use the equation ∆Gf º (g) = ∆Gf º (ℓ) + ∆Gº vap, where ∆Gf º (ℓ) is the standard molar Gibbs energy of formation of the liquid (given as -155.7 kJ/mol), and ∆Gº vap is the standard molar Gibbs energy of vaporization.
Since the vapor pressure of acetone, CH3COCH3 (ℓ), is given in Torr, convert it to Pascal by multiplying by 133.32:
P = 232 Torr * (133.32 Pa/Torr)
Now, you can use the equation:
∆Gº vap = -RT ln(P)
Convert the temperature from Celsius to Kelvin by adding 273.15:
T = 25 ºC + 273.15 = 298.15 K
Plug in the values into the equation:
∆Gº vap = - (8.314 J/mol·K) * 298.15 K * ln(P)
Calculate ∆Gf º (g) by adding ∆Gf º (ℓ) and ∆Gº vap. The unit for both ∆Gf º (ℓ) and ∆Gº vap should be in J/mol.
c) To calculate ∆G for the reaction 2 CuBr2 (s) → 2 CuBr (s) + Br2 (g) at 298 K, you can use the equation ∆G = ∆Gº + RT ln(Q), where ∆Gº is the standard Gibbs energy change of the reaction, R is the gas constant (8.314 J/mol·K), T is the temperature in Kelvin, and Q is the reaction quotient.
First, convert the given equilibrium vapour pressure of Br2 (g) from kPa to Pascal by multiplying by 1000:
P = 1.90 x 10^-6 kPa * 1000 Pa/kPa
Next, calculate Q by dividing the given pressure of Br2 (g) (1.00 x 10^-7 kPa) by the equilibrium vapour pressure of Br2 (g) (P).
Now, you can use the equation:
∆G = ∆Gº + RT ln(Q)
Plug in the values into the equation:
∆G = ∆Gº + (8.314 J/mol·K) * 298 K * ln(Q)
This will give you the value of ∆G in J/mol.
d) To determine whether nitrous acid will spontaneously dissociate in the given conditions, you can compare the reaction quotient (Q) to the equilibrium constant (K). If Q > K, the reaction proceeds in the forward direction and is spontaneous. If Q < K, the reaction proceeds in the reverse direction and is not spontaneous.
For the reaction HNO2 (aq) + H2O (ℓ) ↔ NO2¯ (g) + H3O+ (aq), the equilibrium constant K is given as 4.3 x 10^-4 at 25 ºC.
a) When [HNO2] = [NO2¯] = [H3O+] = 1.0 M, calculate Q by substituting these concentrations into the expression for Q.
b) When [HNO2] = 1.0 M and [NO2¯] = [H3O+] = 1.0 x 10^-5 M, calculate Q again.
Compare the values of Q to the equilibrium constant K. If Q > K, the reaction is spontaneous and nitrous acid will dissociate. If Q < K, the reaction is not spontaneous and nitrous acid will not dissociate.