Find the temperature at which the reaction below is spontaneous given this data:
Hfo of NH3(g) = - 46 kJ/mol, So of NH3(g) = 192.5 J/mol·K, So of N2(g) = 191.5 J/mol·K,
So of H2(g) = 130.6 J/mol·K.
N2(g) + 3H2(g) ßà 2NH3(g)
To determine the temperature at which the reaction is spontaneous, we can use the Gibbs Free Energy equation:
ΔG = ΔH - TΔS
where ΔG is the Gibbs Free Energy change, ΔH is the enthalpy change, T is the temperature in Kelvin, and ΔS is the entropy change.
In this case, we need to calculate the Gibbs Free Energy change at the temperature where the reaction is spontaneous. To do this, we need to find the values for ΔH and ΔS:
ΔH = ΣnHfo(products) - ΣnHfo(reactants)
ΔS = ΣnSo(products) - ΣnSo(reactants)
Given:
Hfo of NH3(g) = -46 kJ/mol
So of NH3(g) = 192.5 J/mol·K
So of N2(g) = 191.5 J/mol·K
So of H2(g) = 130.6 J/mol·K
We can now substitute these values into the equations:
ΔH = [2(-46 kJ/mol)] - [1(0 kJ/mol) + 3(0 kJ/mol)]
= -92 kJ/mol
ΔS = [2(192.5 J/mol·K)] - [1(191.5 J/mol·K) + 3(130.6 J/mol·K)]
= 193 J/mol·K
Now we can rewrite the Gibbs Free Energy equation:
ΔG = -92 kJ/mol - T(193 J/mol·K)
To calculate the temperature at which the reaction is spontaneous, we can set ΔG equal to zero:
0 = -92 kJ/mol - T(193 J/mol·K)
Solving for T:
T(193 J/mol·K) = -92 kJ/mol
T = (-92 kJ/mol) / (193 J/mol·K)
T = -0.477 kJ/K
Since the temperature cannot be negative, the reaction is spontaneous at a temperature greater than 0 Kelvin.
To determine the temperature at which the reaction is spontaneous, we need to calculate the Gibbs free energy change (ΔG) for the reaction. The equation for ΔG is as follows:
ΔG = ΔH - TΔS
where ΔH is the change in enthalpy, T is the temperature in Kelvin, and ΔS is the change in entropy.
Given that the reaction is spontaneous, we know that ΔG must be negative. Therefore, we can rearrange the equation as follows:
TΔS = ΔH - ΔG
We need to calculate ΔH and ΔS first, then substitute the values into the rearranged equation.
ΔH = ∑Hf(products) - ∑Hf(reactants)
Given that Hfo of NH3(g) = -46 kJ/mol, we can determine ∑Hf(products):
∑Hf(products) = Hfo of NH3(g) = -46 kJ/mol
Since there are two moles of NH3 in the balanced equation, we multiply ∑Hf(products) by 2:
∑Hf(products) = -46 kJ/mol × 2 = -92 kJ/mol
To calculate ∑Hf(reactants), we need to add up the enthalpies of N2(g) and 3H2(g):
∑Hf(reactants) = Hfo of N2(g) + 3 × Hfo of H2(g)
Given that Hfo of N2(g) = 0 kJ/mol and Hfo of H2(g) = 0 kJ/mol, we have:
∑Hf(reactants) = 0 kJ/mol + 3 × 0 kJ/mol = 0 kJ/mol
Now, we can calculate ΔH:
ΔH = ∑Hf(products) - ∑Hf(reactants) = -92 kJ/mol - 0 kJ/mol = -92 kJ/mol
Next, we need to calculate ΔS:
ΔS = ∑S(products) - ∑S(reactants)
Given the molar entropies, we can determine ∑S(products) and ∑S(reactants):
∑S(products) = So of NH3(g) = 192.5 J/mol·K
Again, since there are two moles of NH3 in the balanced equation, we multiply ∑S(products) by 2:
∑S(products) = 192.5 J/mol·K × 2 = 385 J/mol·K
∑S(reactants) = So of N2(g) + 3 × So of H2(g)
Given So of N2(g) = 191.5 J/mol·K and So of H2(g) = 130.6 J/mol·K, we have:
∑S(reactants) = 191.5 J/mol·K + 3 × 130.6 J/mol·K = 583.3 J/mol·K
Now, we can calculate ΔS:
ΔS = ∑S(products) - ∑S(reactants) = 385 J/mol·K - 583.3 J/mol·K = -198.3 J/mol·K
Finally, we can substitute ΔH and ΔS into the rearranged equation to find the temperature at which the reaction is spontaneous:
TΔS = ΔH - ΔG
Since the reaction is spontaneous, ΔG must be negative. Therefore, ΔG is equal to zero:
TΔS = ΔH - ΔG
TΔS = ΔH - 0
TΔS = ΔH
Now, solve for T:
T = ΔH / ΔS
T = -92 kJ/mol / (-198.3 J/mol·K)
Note that we need to convert kJ to J:
T = -92,000 J/mol / (-198.3 J/mol·K)
Dividing the numerator by the denominator gives the temperature in Kelvin:
T ≈ 464.2 K
Therefore, the temperature at which the reaction is spontaneous is approximately 464.2 Kelvin.