To solve for the equilibrium concentrations of all species for the reaction:
\[ 2 \text{CO}_2 (g) \rightleftharpoons 2 \text{CO} (g) + \text{O}_2 (g) \]
given \( K = 2.0 \times 10^{-6} , \text{mol/L} \), we will use the ICE table method.
Step 1: Initial Conditions
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Initial concentration of \(\text{CO}_2\): \[ \text{Initial moles of } \text{CO}_2 = 2.0 , \text{mol} \] \[ \text{Volume} = 5.0 , \text{L} \] \[ [\text{CO}_2] = \frac{2.0, \text{mol}}{5.0, \text{L}} = 0.4 , \text{mol/L} \]
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Initial concentrations of \(\text{CO}\) and \(\text{O}_2\): \[ [\text{CO}] = 0 , \text{mol/L}, \quad [\text{O}_2] = 0 , \text{mol/L} \]
Step 2: ICE Table Setup
| Species | Initial (mol/L) | Change (mol/L) | Equilibrium (mol/L) | |-----------|------------------|-------------------|---------------------------| | \(\text{CO}_2\) | 0.4 | \(-2x\) | \(0.4 - 2x\) | | \(\text{CO}\) | 0 | \(+2x\) | \(2x\) | | \(\text{O}_2\) | 0 | \(+x\) | \(x\) |
Step 3: Write the Equilibrium Expression
For the reaction, the equilibrium constant expression is given by:
\[ K = \frac{[\text{CO}]^2 [\text{O}_2]}{[\text{CO}_2]^2} \]
Substituting the equilibrium concentrations we found:
\[ K = \frac{(2x)^2 (x)}{(0.4 - 2x)^2} \]
Step 4: Setting up the Equation
Now we can substitute \(K\) into the expression we derived:
\[ 2.0 \times 10^{-6} = \frac{(2x)^2 (x)}{(0.4 - 2x)^2} \]
Simplifying the left-hand side:
\[ 2.0 \times 10^{-6} = \frac{4x^2 \cdot x}{(0.4 - 2x)^2} \]
This simplifies to:
\[ 2.0 \times 10^{-6} = \frac{4x^3}{(0.4 - 2x)^2} \]
Step 5: Cross-multiplying
Now we cross-multiply to eliminate the fraction:
\[ 2.0 \times 10^{-6} (0.4 - 2x)^2 = 4x^3 \]
Expanding the left-hand side:
\[ 2.0 \times 10^{-6} (0.16 - 1.6x + 4x^2) = 4x^3 \]
Distributing:
\[ 3.2 \times 10^{-7} - 3.2 \times 10^{-6} x + 8.0 \times 10^{-6} x^2 = 4x^3 \]
Rearranging gives:
\[ 4x^3 - 8.0 \times 10^{-6} x^{2} + 3.2 \times 10^{-6} x - 3.2 \times 10^{-7} = 0 \]
Step 6: Solving for \(x\)
This cubic equation can be solved using numerical methods or software. However, for smaller \(K\), we may assume \(x\) is very small compared to the initial concentration of \(CO_2\), so we can use the approximation:
Assume \(0.4 - 2x \approx 0.4\):
\[ 2.0 \times 10^{-6} (0.4)^2 \approx 4x^3 \implies x^3 \approx \frac{2.0 \times 10^{-6} \cdot 0.16}{4} \]
Calculating gives:
\[ x^3 \approx \frac{3.2 \times 10^{-7}}{4} \approx 8.0 \times 10^{-8} \] \[ x \approx (8.0 \times 10^{-8})^{1/3} \approx 4.3 \times 10^{-3} , \text{mol/L} \]
Step 7: Final Concentrations
Now we substitute \(x\) back to find the equilibrium concentrations:
- \([\text{CO}_2] = 0.4 - 2(4.3 \times 10^{-3}) \approx 0.4 - 0.0086 \approx 0.3914 , \text{mol/L}\)
- \([\text{CO}] = 2(4.3 \times 10^{-3}) \approx 0.0086 , \text{mol/L}\)
- \([\text{O}_2] = 4.3 \times 10^{-3} , \text{mol/L}\)
Equilibrium concentrations:
- \([\text{CO}_2] \approx 0.3914 , \text{mol/L}\)
- \([\text{CO}] \approx 0.0086 , \text{mol/L}\)
- \([\text{O}_2] \approx 0.0043 , \text{mol/L}\)
These are the equilibrium concentrations for the reaction under the given conditions.