Question

using any data you can find in the aleks data resource, calculate the equilibrium constant k at 25.0 °c for the following reaction.
fe₂o₃(s) + 3 h₂(g) → 2 fe(s) + 3 h₂o(l)
round your answer to 2 significant digits.
k =
thermodynamic properties of pure substances
i⁻(aq) -55.2 -51.6 111.3
hi(g) 26.5 1.7 206.6
iron
fe(s) 0 0 27.3
fe²⁺(aq) -89.1 -78.9 -137.7
fe³⁺(aq) -48.5 -4.7 -315.9
fecl₂(s) -341.8 -302.3 118.0
fecl₃(s) -399.5 -334.0 142.3
feo(s) -272.0 -255.2 60.75
fe₂o₃(s) -824.2 -742.2 87.4
fe₃o₄(s) -1118.4 -1015.4 146.4
fe(oh)₂(s) -574.04 -483.55 87.93
fe(oh)₃(s) -832.62 -696.5 104.56
lead
pb(s) 0 0 64.8
pb²⁺(aq) -1.7 -24.4 10.5
pbbr₂(s) -278.7 -261.9 161.5
pbcl₂(s) -359.4 -314.1 136.0
pbo(s; litharge) -219.0 -188.9 66.5
pbo(s; massicot) -217.3 -187.9 68.7

Explanation:

Step1: Recall the formula for \( \Delta G^\circ \)

The relationship between the equilibrium constant \( K \) and the standard Gibbs free energy change \( \Delta G^\circ \) is \( \Delta G^\circ = -RT\ln K \), where \( R = 8.314 \, \text{J/(mol·K)} \) and \( T = 25.0^\circ \text{C} = 298.15 \, \text{K} \). First, we need to calculate \( \Delta G^\circ \) for the reaction using the formula \( \Delta G^\circ = \sum \Delta G_f^\circ(\text{products}) - \sum \Delta G_f^\circ(\text{reactants}) \).

Step2: Identify \( \Delta G_f^\circ \) values

From the data:
- \( \Delta G_f^\circ(\text{Fe}_2\text{O}_3(s)) = -742.2 \, \text{kJ/mol} \)
- \( \Delta G_f^\circ(\text{H}_2(g)) = 0 \, \text{kJ/mol} \) (standard state for elements)
- \( \Delta G_f^\circ(\text{Fe}(s)) = 0 \, \text{kJ/mol} \) (standard state for elements)
- \( \Delta G_f^\circ(\text{H}_2\text{O}(l)) = -237.13 \, \text{kJ/mol} \) (we need this value, assume it's a known standard value since it's not in the given table but is a standard thermodynamic value)

Step3: Calculate \( \sum \Delta G_f^\circ(\text{products}) \)

Products: \( 2\text{Fe}(s) + 3\text{H}_2\text{O}(l) \)
\( \sum \Delta G_f^\circ(\text{products}) = 2 \times \Delta G_f^\circ(\text{Fe}(s)) + 3 \times \Delta G_f^\circ(\text{H}_2\text{O}(l)) = 2 \times 0 + 3 \times (-237.13) = -711.39 \, \text{kJ/mol} \)

Step4: Calculate \( \sum \Delta G_f^\circ(\text{reactants}) \)

Reactants: \( \text{Fe}_2\text{O}_3(s) + 3\text{H}_2(g) \)
\( \sum \Delta G_f^\circ(\text{reactants}) = \Delta G_f^\circ(\text{Fe}_2\text{O}_3(s)) + 3 \times \Delta G_f^\circ(\text{H}_2(g)) = -742.2 + 3 \times 0 = -742.2 \, \text{kJ/mol} \)

Step5: Calculate \( \Delta G^\circ \)

\( \Delta G^\circ = \sum \Delta G_f^\circ(\text{products}) - \sum \Delta G_f^\circ(\text{reactants}) = (-711.39) - (-742.2) = 30.81 \, \text{kJ/mol} = 30810 \, \text{J/mol} \) (converted to J/mol for units with \( R \))

Step6: Solve for \( K \) from \( \Delta G^\circ = -RT\ln K \)

Rearrange the formula: \( \ln K = \frac{-\Delta G^\circ}{RT} \)
Substitute values: \( \ln K = \frac{-30810}{8.314 \times 298.15} \approx \frac{-30810}{2479.0} \approx -12.43 \)
Then, \( K = e^{-12.43} \approx 3.7 \times 10^{-6} \)? Wait, no, wait, maybe I made a mistake in \( \Delta G_f^\circ(\text{H}_2\text{O}(l)) \). Wait, let's check again. Wait, maybe the given table has some missing values, but actually, let's re - evaluate. Wait, the reaction is \( \text{Fe}_2\text{O}_3(s)+3\text{H}_2(g) ightarrow 2\text{Fe}(s)+3\text{H}_2\text{O}(l) \). Let's use the correct \( \Delta G_f^\circ \) for \( \text{H}_2\text{O}(l) \) which is \( -237.13 \, \text{kJ/mol} \).

Wait, maybe I messed up the sign. Let's recalculate \( \Delta G^\circ \):

\( \Delta G^\circ = [2\Delta G_f^\circ(\text{Fe}(s)) + 3\Delta G_f^\circ(\text{H}_2\text{O}(l))] - [\Delta G_f^\circ(\text{Fe}_2\text{O}_3(s)) + 3\Delta G_f^\circ(\text{H}_2(g))] \)

\( = [2\times0 + 3\times(-237.13)] - [-742.2 + 3\times0] \)

\( = (-711.39) - (-742.2) = 30.81 \, \text{kJ/mol} = 30810 \, \text{J/mol} \)

Now, \( \ln K=\frac{-\Delta G^\circ}{RT}=\frac{-30810}{8.314\times298.15}\approx\frac{-30810}{2479}\approx - 12.43 \)

\( K = e^{-12.43}\approx3.7\times 10^{-6} \)? Wait, that can't be right. Wait, maybe I used the wrong \( \Delta G_f^\circ \) for \( \text{Fe}_2\text{O}_3 \). Wait, the table shows \( \Delta G_f^\circ(\text{Fe}_2\text{O}_3(s))=-742.2 \, \text{kJ/mol} \), \( \Delta G_f^\circ(\text{H}_2\text{O}(l)) \) is actually \( -237.1 \, \text{kJ/mol} \) (more accurately). Let's recalculate:

\( \sum \text{products: } 2\times0 + 3\times(-237.1)= - 711.3 \, \text{kJ/mol} \)

\( \sum \text{reactants: }-742.2+3\times0=-742.2 \, \text{kJ/mol} \)

\( \Delta G^\circ=-711.3 - (-742.2)=30.9 \, \text{kJ/mol}=30900 \, \text{J/mol} \)

\( \ln K=\frac{-30900}{8.314\times298.15}=\frac{-30900}{2479}\approx - 12.46 \)

\( K = e^{-12.46}\approx3.5\times 10^{-6} \)? Wait, no, maybe I have the reaction reversed? Wait, no, the reaction is \( \text{Fe}_2\text{O}_3 + \text{H}_2 ightarrow\text{Fe}+\text{H}_2\text{O} \), which is a reduction of iron oxide by hydrogen. Wait, maybe the \( \Delta G_f^\circ(\text{H}_2\text{O}(l)) \) is actually \( -237.1 \, \text{kJ/mol} \), but let's check the correct approach.

Wait, another way: Maybe the given table has some errors, but let's assume that we can get \( \Delta G^\circ \) correctly. Wait, perhaps I made a mistake in the sign. Wait, \( \Delta G^\circ=\sum \text{products} - \sum \text{reactants} \). So products are \( 2\text{Fe}(s) \) ( \( \Delta G_f = 0 \)) and \( 3\text{H}_2\text{O}(l) \) ( \( \Delta G_f=-237.1 \, \text{kJ/mol} \) ), reactants are \( \text{Fe}_2\text{O}_3(s) \) ( \( \Delta G_f=-742.2 \, \text{kJ/mol} \)) and \( 3\text{H}_2(g) \) ( \( \Delta G_f = 0 \) ). So:

\( \Delta G^\circ=(2\times0 + 3\times(-237.1))-(-742.2 + 3\times0)=-711.3 + 742.2 = 30.9 \, \text{kJ/mol}=30900 \, \text{J/mol} \)

Now, \( \ln K=\frac{-\Delta G^\circ}{RT}=\frac{-30900}{8.314\times298.15}\approx - 12.46 \)

\( K = e^{-12.46}\approx3.5\times 10^{-6} \)? But this seems small. Wait, maybe the \( \Delta G_f^\circ(\text{H}_2\text{O}(l)) \) is \( -285.8 \, \text{kJ/mol} \)? No, that's for enthalpy. Wait, no, \( \Delta G_f^\circ(\text{H}_2\text{O}(l))=-237.13 \, \text{kJ/mol} \), \( \Delta H_f^\circ(\text{H}_2\text{O}(l))=-285.83 \, \text{kJ/mol} \). I think I confused them earlier.

Wait, let's start over. The correct formula for \( \Delta G^\circ \) calculation:

For the reaction \( aA + bB ightarrow cC + dD \), \( \Delta G^\circ = c\Delta G_f^\circ(C)+d\Delta G_f^\circ(D)-a\Delta G_f^\circ(A)-b\Delta G_f^\circ(B) \)

Reaction: \( \text{Fe}_2\text{O}_3(s)+3\text{H}_2(g) ightarrow 2\text{Fe}(s)+3\text{H}_2\text{O}(l) \)

So:

\( \Delta G^\circ = [2\Delta G_f^\circ(\text{Fe}(s)) + 3\Delta G_f^\circ(\text{H}_2\text{O}(l))] - [\Delta G_f^\circ(\text{Fe}_2\text{O}_3(s)) + 3\Delta G_f^\circ(\text{H}_2(g))] \)

We know:

- \( \Delta G_f^\circ(\text{Fe}(s)) = 0 \, \text{kJ/mol} \)
- \( \Delta G_f^\circ(\text{H}_2(g)) = 0 \, \text{kJ/mol} \)
- \( \Delta G_f^\circ(\text{Fe}_2\text{O}_3(s))=-742.2 \, \text{kJ/mol} \) (from table)
- \( \Delta G_f^\circ(\text{H}_2\text{O}(l))=-237.13 \, \text{kJ/mol} \) (standard value)

Substitute:

\( \Delta G^\circ = [2\times0 + 3\times(-237.13)] - [-742.2 + 3\times0] \)

\( = (-711.39) - (-742.2) \)

\( = 30.81 \, \text{kJ/mol}=30810 \, \text{J/mol} \)

Now, use \( \Delta G^\circ=-RT\ln K \)

\( \ln K=\frac{-\Delta G^\circ}{RT}=\frac{-30810}{8.314\times298.15} \)

Calculate \( 8.314\times298.15 = 8.314\times298 + 8.314\times0.15\approx2477.57 + 1.247\approx2478.82 \)

\( \ln K=\frac{-30810}{2478.82}\approx - 12.43 \)

\( K = e^{-12.43}\approx3.7\times 10^{-6} \)? But this seems incorrect. Wait, maybe the reaction is actually \( \text{Fe}_2\text{O}_3(s)+3\text{H}_2(g) ightarrow 2\text{Fe}(s)+3\text{H}_2\text{O}(g) \), but the problem says \( \text{H}_2\text{O}(l) \). Wait, no, the problem states \( \text{H}_2\text{O}(l) \).

Wait, maybe I made a mistake in the sign of \( \Delta G^\circ \). Let's check the reaction spontaneity. If \( \Delta G^\circ \) is positive, \( K \) is less than 1, which would mean the reaction is non - spontaneous under standard conditions, which makes sense for this reaction? Wait, no, the reduction of iron oxide by hydrogen is spontaneous at high temperatures, but at 25°C, maybe it's non - spontaneous.

Wait, let's recalculate \( \Delta G^\circ \) with the correct \( \Delta G_f^\circ(\text{H}_2\text{O}(l))=-237.13 \, \text{kJ/mol} \):

\( \sum \text{products: } 2\times0+3\times(-237.13)= - 711.39 \, \text{kJ/mol} \)

\( \sum \text{reactants: }-742.2 + 3\times0=-742.2 \, \text{kJ/mol} \)

\( \Delta G^\circ=-711.39-(-742.2)=30.81 \, \text{kJ/mol} \)

Now, \( \ln K=\frac{-30810}{8.314\times298.15}\approx - 12.43 \)

\( K = e^{-12.43}\approx3.7\times 10^{-6} \). But when we round to two significant digits, it's \( 4\times 10^{-6} \)? Wait, no, \( e^{-12.43}\approx3.7\times 10^{-6}\approx4\times 10^{-6} \) (two significant digits). Wait, but maybe I have the wrong \( \Delta G_f^\circ \) for \( \text{H}_2\text{O}(l) \). Let's check a standard table: \( \Delta G_f^\circ(\text{H}_2\text{O}(l))=-237.13 \, \text{kJ/mol} \) is correct.

Wait, another approach: Maybe the given table has \( \Delta G_f^\circ \) values in kJ/mol, and we can use the formula correctly. So the steps are:

1. Calculate \( \Delta G^\circ \) using \( \Delta G^\circ=\sum \Delta G_f^\circ(\text{products})-\sum \Delta G_f^\circ(\text{reactants}) \)
2. Convert \( \Delta G^\circ \) to J/mol.
3. Use \( \Delta G^\circ=-RT\ln K \) to solve for \( K \).

After performing the calculations, we get \( K\approx4\times 10^{-6} \) (rounded to two significant digits). Wait, but let's do the calculation more accurately.

\( \ln K=\frac{-30810}{8.314\times298.15}=\frac{-30810}{2478.8}\approx - 12.43 \)

\( K = e^{-12.43}\approx e^{-12}\times e^{-0.43}\approx6.14\times 10^{-6}\times0.651\approx3.99\times 10^{-6}\approx4.0\times 10^{-6} \)

Answer:

\( \boxed{4.0\times 10^{-6}} \) (Note: The answer may vary slightly depending on the exact value of \( \Delta G_f^\circ(\text{H}_2\text{O}(l)) \) used, but with two significant digits, it is approximately \( 4\times 10^{-6} \))