WO3 (s) + 3 H2 (g) -> W (s) + 3 H20 (g)
Tungsten is obtained commercially by the reduction of WO3 with hydrogen according to the equation above. The following data related to this reaction are available.
DeltaH(kilojoules/mole) for WO3 is
-839.5. DeltaH for H20(g) is -241.6.
Delta G for WO3 is -762.7. Delta G for H20 i -228.2
A) What is the value of the equilibrium constant for the system represented above?
B) Calculate Delta S at 25C for the reaction indicated above
A) First, we need to calculate the standard Gibbs free energy change (ΔG°) for the reaction.
ΔG° = ΣΔG°(products) - ΣΔG°(reactants)
ΔG° = [1 × ΔG(W) + 3 × ΔG(H2O)] - [ΔG(WO3) + 3 × ΔG(H2)]
Using the given values:
ΔG° = [1 × 0 + 3 × (-228.2)] - [(-762.7) + 3 × 0] (Note: ΔG° of W or its Grubbs free energy change, since it is a pure element, is zero)
ΔG° = -684.6 J/mol
Now, we can calculate the equilibrium constant (K) using ΔG° and the following equation:
ΔG° = -RT ln K
where R is the gas constant (8.314 J/mol K) and T is the temperature in Kelvin (25°C = 298.15 K)
Rearrange the equation and solve for K:
ln K = -ΔG° / RT
K = exp(-ΔG° / RT)
K = exp(-(-684.6) / (8.314 × 298.15))
K ≈ 2355.6
The value of the equilibrium constant for the system is approximately 2355.6.
B) We can calculate the standard entropy change (ΔS°) using ΔH° and ΔG° with the following equation:
ΔG° = ΔH° - TΔS°
First, we need to calculate the standard enthalpy change (ΔH°) for the reaction.
ΔH° = ΣΔH°(products) - ΣΔH°(reactants)
ΔH° = [1 × ΔH(W) + 3 × ΔH(H2O)] - [ΔH(WO3) + 3 × ΔH(H2)]
Using the given values:
ΔH° = [1 × 0 + 3 × (-241.6)] - [(-839.5) + 3 × 0] (Note: ΔH° of W, since it is apure element, is zero)
ΔH° = -88.3 kJ/mol
Now, we can solve for ΔS° using ΔG° and ΔH°:
ΔS° = (ΔH° - ΔG°) / T
ΔS° = ((-88.3 × 1000) - (-684.6)) / 298.15 (Note: Convert ΔH° to J/mol by multiplying by 1000)
ΔS° ≈ -294.9 J/mol K
The change in entropy ΔS at 25°C for the reaction is approximately -294.9 J/mol K.
To calculate the equilibrium constant (K) for the given reaction, we need to use the equation:
ΔG = -RTln(K)
where ΔG is the change in Gibbs free energy, R is the gas constant (8.314 J/(mol·K)), T is the temperature in Kelvin, and ln is the natural logarithm.
A) To find the equilibrium constant:
ΔG = -762.7 kJ/mol
Convert ΔG to J/mol:
ΔG = -762.7 × 1000 J/mol = -762,700 J/mol
ΔG for H2O(g) = -228.2 kJ/mol
Convert ΔG to J/mol:
ΔG = -228.2 × 1000 J/mol = -228,200 J/mol
Sum of ΔG values:
ΔG(sum) = ΔG(WO3) + 3 × ΔG(H2) - ΔG(W) - 3 × ΔG(H2O)
ΔG(sum) = -762,700 J/mol + 3 × 0 J/mol - 0 J/mol - 3 × -228,200 J/mol
ΔG(sum) = -762,700 J/mol + 684,600 J/mol + 684,600 J/mol
ΔG(sum) = 606,500 J/mol
Now, convert ΔG to kJ/mol:
ΔG = 606.5 kJ/mol
Since T (temperature) is not provided, we assume it to be 298 K (25°C).
R = 8.314 J/(mol·K)
ΔG = 606.5 kJ/mol = 606,500 J/mol
T = 298 K
Now, let's solve the equation for K:
ΔG = -RTln(K)
606,500 J/mol = -(8.314 J/(mol·K)) × 298 K × ln(K)
Rearrange the equation to solve for ln(K):
ln(K) = -(606,500 J/mol) / (8.314 J/(mol·K) × 298 K)
ln(K) = -235.27
Now, calculate K:
K = e^ln(K)
K = e^-235.27
K ≈ 8.195 × 10^-103 (rounded to the appropriate significant figures)
So, the value of the equilibrium constant for the system represented above is approximately 8.195 × 10^-103.
B) To calculate ΔS (change in entropy):
ΔH(WO3) = -839.5 kJ/mol
ΔH(H2O(g)) = -241.6 kJ/mol
Sum of ΔH values:
ΔH(sum) = ΔH(WO3) + 3 × 0 kJ/mol - 0 kJ/mol - 3 × ΔH(H2O)
ΔH(sum) = -839.5 kJ/mol - 3 × (-241.6 kJ/mol)
ΔH(sum) = -839.5 kJ/mol + 724.8 kJ/mol
ΔH(sum) = -114.7 kJ/mol
Now, ΔS can be calculated using the equation:
ΔG = ΔH - TΔS
ΔG = -762.7 kJ/mol
ΔH = -114.7 kJ/mol
T = 298 K
ΔG = ΔH - TΔS
-762.7 kJ/mol = -114.7 kJ/mol - (298 K)(-ΔS)
-762.7 kJ/mol + 114.7 kJ/mol = 298 K × ΔS
-648 kJ/mol = 298 K × ΔS
Now, solve for ΔS:
ΔS = (-648 kJ/mol) / (298 K)
ΔS ≈ -2.18 kJ/(mol·K) (rounded to the appropriate significant figures)
Therefore, ΔS at 25°C for the reaction indicated above is approximately -2.18 kJ/(mol·K).
To answer these questions, we can use the relationship between the equilibrium constant (K) and the standard Gibbs free energy change (ΔG°) at a given temperature (T):
ΔG° = -RT ln(K)
Where:
ΔG° is the standard Gibbs free energy change for the reaction
R is the gas constant (8.314 J/(mol·K) or 0.008314 kJ/(mol·K))
T is the temperature in Kelvin
ln denotes the natural logarithm
A) What is the value of the equilibrium constant for the system represented above?
We are given the ΔG° values for WO3 and H2O, but not the ΔG° for hydrogen gas (H2). However, we can calculate it using the equation:
ΔG°(reaction) = ΣnΔG°(products) - ΣmΔG°(reactants)
For the given reaction:
ΔG°(reaction) = ΔG°(W) + 3ΔG°(H2O) - ΔG°(WO3) - 3ΔG°(H2)
Substituting the given values:
ΔG°(reaction) = 0 - 3(-228.2) - (-762.7) - 3ΔG°(H2)
Simplifying the equation:
ΔG°(H2) = -ΔG°(reaction) + 3(-228.2) - 762.7
ΔG°(H2) = -ΔG°(reaction) - 684.1
Now that we have the ΔG° of H2, we can calculate the equilibrium constant (K) at the given temperature (25°C = 298 K):
ΔG° = -RT ln(K)
-ΔG°(reaction) - 684.1 = -8.314 × 298 × ln(K)
Solving for K:
ln(K) = (ΔG°(reaction) + 684.1) / (8.314 × 298)
K = e^[(ΔG°(reaction) + 684.1) / (8.314 × 298)]
This will give you the value of the equilibrium constant (K) for the system represented above.
B) Calculate ΔS at 25°C for the reaction indicated above
To calculate ΔS (change in entropy) at a specific temperature, we can use the equation:
ΔG° = ΔH° - TΔS°
Where:
ΔG° is the standard Gibbs free energy change for the reaction
ΔH° is the standard enthalpy change for the reaction
T is the temperature in Kelvin
ΔS° is the standard entropy change for the reaction
We are given the ΔH° values for WO3 and H2O, but not the ΔS° values. However, we can calculate them using the equation:
ΔG° = ΔH° - TΔS°
For the given reaction:
ΔG° = -762.7 kJ/mol
ΔH° = -839.5 kJ/mol
T = 298 K (25°C)
Solving for ΔS°:
ΔG° = ΔH° - TΔS°
-762.7 = -839.5 - (298 × ΔS°)
Simplifying the equation:
ΔS° = (-762.7 + 839.5) / 298
This will give you the value of ΔS° at 25°C for the reaction indicated above.