Question

using any data you can find in the aleks data resource, calculate the equilibrium constant ( k ) at ( 25.0^circ \text{c} ) for the following reaction.
( \text{sno}_2(s) + 2\text{co}(g)
ightarrow \text{sn}(s, \text{white}) + 2\text{co}_2(g) )
round your answer to 2 significant digits.
( k = square )
thermodynamic properties of pure substances
( \text{h}_2\text{s(aq)} ) -39.0 -27.4 122.0
( \text{so(g)} ) 6.3 -19.9 222.0
( \text{so}_2\text{(g)} ) -296.8 -300.1 248.2
( \text{so}_3\text{(g)} ) -395.7 -371.1 256.8
( \text{so}_3^{2-}\text{(aq)} ) -635.5 -486.5 -29.0
( \text{so}_4^{2-}\text{(aq)} ) -909.3 -744.5 20.1
( \text{hso}_3^-\text{(aq)} ) -626.2 -527.7 139.7
( \text{hso}_4^-\text{(aq)} ) -887.3 -755.9 131.8
( \text{h}_2\text{so}_4\text{(l)} ) -814.0 -690.0 156.9
( \text{h}_2\text{so}_4\text{(aq)} ) -909.3 -744.5 20.1
tin
( \text{sn(s; white)} ) 0 0 51.2
( \text{sn(s; gray)} ) -2.1 0.1 44.1
( \text{sncl}_4\text{(l)} ) -511.3 -440.1 258.6
( \text{sno}_2\text{(s)} ) -577.6 -515.8 49.0
titanium
( \text{ti(s)} ) 0 0 30.7
( \text{ticl}_4\text{(l)} ) -804.2 -737.2 252.3
( \text{ticl}_4\text{(g)} ) -763.2 -726.3 353.2

Explanation:

Step1: Find ΔG° for the reaction

First, we need the standard Gibbs free energy of formation ($\Delta G_f^\circ$) values for each compound. From the table:
- $\Delta G_f^\circ(\text{SnO}_2(s)) = -515.8\ \text{kJ/mol}$
- $\Delta G_f^\circ(\text{CO}(g))$ (we need this, but it's not in the shown table; typically, $\Delta G_f^\circ(\text{CO}(g)) = -137.2\ \text{kJ/mol}$)
- $\Delta G_f^\circ(\text{Sn}(s, \text{white})) = 0\ \text{kJ/mol}$
- $\Delta G_f^\circ(\text{CO}_2(g)) = -394.4\ \text{kJ/mol}$ (standard value, since it's not fully shown here, but standard data: $\Delta G_f^\circ(\text{CO}_2) = -394.4\ \text{kJ/mol}$)

The reaction is: $\text{SnO}_2(s) + 2\ \text{CO}(g)
ightarrow \text{Sn}(s, \text{white}) + 2\ \text{CO}_2(g)$

Using the formula for $\Delta G^\circ_{\text{rxn}} = \sum \Delta G_f^\circ(\text{products}) - \sum \Delta G_f^\circ(\text{reactants})$

$\sum \Delta G_f^\circ(\text{products}) = \Delta G_f^\circ(\text{Sn}) + 2\Delta G_f^\circ(\text{CO}_2) = 0 + 2(-394.4) = -788.8\ \text{kJ/mol}$

$\sum \Delta G_f^\circ(\text{reactants}) = \Delta G_f^\circ(\text{SnO}_2) + 2\Delta G_f^\circ(\text{CO}) = -515.8 + 2(-137.2) = -515.8 - 274.4 = -790.2\ \text{kJ/mol}$

$\Delta G^\circ_{\text{rxn}} = (-788.8) - (-790.2) = 1.4\ \text{kJ/mol}$ (Wait, this seems off. Wait, maybe the table has $\text{CO}_2$? Wait, the shown table has $\text{SO}_2, \text{SO}_3$, but not $\text{CO}$ or $\text{CO}_2$. Wait, maybe I made a mistake. Wait, the user's table shows part of it. Let's check again. Wait, the table given has $\text{SnO}_2(s)$: -515.8, $\text{Sn}(s, white)$: 0. Let's use standard values for $\text{CO}$ and $\text{CO}_2$ (since they are not in the shown table, but standard thermo data):

$\Delta G_f^\circ(\text{CO}(g)) = -137.2\ \text{kJ/mol}$, $\Delta G_f^\circ(\text{CO}_2(g)) = -394.4\ \text{kJ/mol}$

So:

$\Delta G^\circ_{\text{rxn}} = [\Delta G_f^\circ(\text{Sn}) + 2\Delta G_f^\circ(\text{CO}_2)] - [\Delta G_f^\circ(\text{SnO}_2) + 2\Delta G_f^\circ(\text{CO})]$

Plugging in:

$[0 + 2(-394.4)] - [-515.8 + 2(-137.2)]$

$= (-788.8) - [-515.8 - 274.4]$

$= (-788.8) - (-790.2)$

$= 1.4\ \text{kJ/mol} = 1400\ \text{J/mol}$ (convert to J for $R$ units)

Step2: Relate ΔG° to K using $\Delta G^\circ = -RT\ln K$

$R = 8.314\ \text{J/(mol·K)}$, $T = 25.0^\circ\text{C} = 298.15\ \text{K}$

Rearranging: $\ln K = -\frac{\Delta G^\circ}{RT}$

Plugging in:

$\ln K = -\frac{1400\ \text{J/mol}}{(8.314\ \text{J/(mol·K)})(298.15\ \text{K})}$

Calculate denominator: $8.314 \times 298.15 \approx 2479$

$\ln K \approx -\frac{1400}{2479} \approx -0.5647$

Then, $K = e^{-0.5647} \approx 0.56$

Wait, but maybe the $\Delta G_f^\circ(\text{CO})$ from standard data is correct? Wait, maybe I used the wrong $\Delta G_f^\circ(\text{CO})$. Wait, standard $\Delta G_f^\circ(\text{CO}) = -137.2\ \text{kJ/mol}$, $\Delta G_f^\circ(\text{CO}_2) = -394.4\ \text{kJ/mol}$. Let's recalculate:

$\sum \text{products: } 0 + 2(-394.4) = -788.8$

$\sum \text{reactants: } -515.8 + 2(-137.2) = -515.8 - 274.4 = -790.2$

$\Delta G^\circ = -788.8 - (-790.2) = 1.4\ \text{kJ/mol} = 1400\ \text{J/mol}$

Then $\ln K = -1400/(8.314*298.15) \approx -1400/2479 \approx -0.5647$

$K = e^{-0.5647} \approx 0.56$

But wait, maybe the table has $\Delta G_f^\circ(\text{CO})$? Wait, the user's table doesn't show CO, so we use standard values. Alternatively, maybe the $\Delta G_f^\circ(\text{CO})$ is different? Wait, no, standard is -137.2.

Wait, maybe I made a mistake in the reaction. Wait, the reaction is SnO2 + 2 CO → Sn + 2 CO2. Let's check with correct standard values:

$\Delta G_f^\circ(\text{SnO}_2) = -515.8\ \text{kJ/mol}$ (from table)

$\Delta G_f^\circ(\text{CO}) = -137.2\ \text{kJ/mol}$ (standard)

$\Delta G_f^\circ(\text{CO}_2) = -394.4\ \text{kJ/mol}$ (standard)

$\Delta G_f^\circ(\text{Sn}) = 0$

So:

$\Delta G^\circ = [0 + 2(-394.4)] - [-515.8 + 2(-137.2)]$

$= (-788.8) - (-515.8 - 274.4)$

$= (-788.8) - (-790.2)$

$= 1.4\ \text{kJ/mol}$

Then $K = e^{-\Delta G^\circ/(RT)} = e^{-1400/(8.314*298)} \approx e^{-0.565} \approx 0.56$

But maybe the actual calculation with correct $\Delta G_f^\circ(\text{CO})$ (if table had it, but since it's not shown, we use standard). Alternatively, maybe the problem expects using the given table, but the table doesn't have CO or CO2, so we use standard values.

Wait, maybe I messed up the sign. Let's re-express $\Delta G^\circ_{\text{rxn}} = \sum \Delta G_f^\circ(\text{products}) - \sum \Delta G_f^\circ(\text{reactants})$

Products: Sn (0) + 2 CO2 (2*(-394.4)) = -788.8

Reactants: SnO2 (-515.8) + 2 CO (2*(-137.2)) = -515.8 - 274.4 = -790.2

So $\Delta G^\circ = -788.8 - (-790.2) = 1.4\ \text{kJ/mol}$ (positive, so K < 1)

Then $\ln K = -1400/(8.314*298) ≈ -0.565$, so K ≈ 0.56

But maybe the correct standard $\Delta G_f^\circ(\text{CO})$ is -137.168 kJ/mol, $\Delta G_f^\circ(\text{CO}_2)$ is -394.359 kJ/mol. Let's recalculate with more precision:

$\sum$ products: 0 + 2*(-394.359) = -788.718

$\sum$ reactants: -515.8 + 2*(-137.168) = -515.8 - 274.336 = -790.136

$\Delta G^\circ = -788.718 - (-790.136) = 1.418\ \text{kJ/mol} = 1418\ \text{J/mol}$

$\ln K = -1418/(8.314*298.15) = -1418/(2479.0) ≈ -0.572$

$K = e^{-0.572} ≈ 0.564 ≈ 0.56$ (to 2 sig figs)

Answer:

$\boxed{0.56}$ (or around 0.56, depending on precise $\Delta G_f^\circ$ values)