1. **State the problem:** We have the equilibrium reaction $$\text{CO}(g) + \text{H}_2\text{O}(g) \rightleftharpoons \text{H}_2(g) + \text{CO}_2(g)$$ with initial equilibrium moles: CO = 0.4, H2O = 0.3, H2 = 0.6, CO2 = 0.2. We want to find how much CO2 must be added so that the new equilibrium moles of CO2 become 0.6. 2. **Write the equilibrium expression:** $$Q = \frac{[\text{H}_2][\text{CO}_2]}{[\text{CO}][\text{H}_2\text{O}]}$$ 3. **Define variables:** Let $x$ be the moles of CO2 added initially. After adding $x$, the system will shift to re-establish equilibrium. Let the change in moles due to shift be $y$. 4. **Set up expressions for new equilibrium moles:** - CO2 initial after addition: $0.2 + x$ - CO2 at new equilibrium: $0.6$ (given) - Change in CO2 due to shift: $y = 0.6 - (0.2 + x) = 0.4 - x$ Since CO2 is produced in the forward reaction, an increase in CO2 will shift equilibrium to the left, consuming CO2 and H2 and producing CO and H2O. 5. **Write new equilibrium moles for all species:** - CO2: $0.2 + x - y = 0.6$ - H2: $0.6 - y$ - CO: $0.4 + y$ - H2O: $0.3 + y$ 6. **Use the equilibrium constant $K$ from initial conditions:** Calculate $K$ using initial equilibrium moles: $$K = \frac{0.6 \times 0.2}{0.4 \times 0.3} = \frac{0.12}{0.12} = 1$$ 7. **Apply $K$ to new equilibrium:** $$1 = \frac{(0.6 - y)(0.2 + x - y)}{(0.4 + y)(0.3 + y)}$$ Substitute $y = 0.4 - x$: $$1 = \frac{(0.6 - (0.4 - x))(0.2 + x - (0.4 - x))}{(0.4 + (0.4 - x))(0.3 + (0.4 - x))}$$ Simplify numerator: $$0.6 - 0.4 + x = 0.2 + x$$ $$0.2 + x - 0.4 + x = (0.2 - 0.4) + 2x = -0.2 + 2x$$ Simplify denominator: $$0.4 + 0.4 - x = 0.8 - x$$ $$0.3 + 0.4 - x = 0.7 - x$$ So: $$1 = \frac{(0.2 + x)(-0.2 + 2x)}{(0.8 - x)(0.7 - x)}$$ 8. **Multiply both sides by denominator:** $$(0.8 - x)(0.7 - x) = (0.2 + x)(-0.2 + 2x)$$ 9. **Expand both sides:** Left: $$0.8 \times 0.7 - 0.8x - 0.7x + x^2 = 0.56 - 1.5x + x^2$$ Right: $$0.2 \times (-0.2) + 0.2 \times 2x + x \times (-0.2) + x \times 2x = -0.04 + 0.4x - 0.2x + 2x^2 = -0.04 + 0.2x + 2x^2$$ 10. **Set equation:** $$0.56 - 1.5x + x^2 = -0.04 + 0.2x + 2x^2$$ 11. **Bring all terms to one side:** $$0.56 + 0.04 - 1.5x - 0.2x + x^2 - 2x^2 = 0$$ $$0.6 - 1.7x - x^2 = 0$$ 12. **Rewrite quadratic:** $$-x^2 - 1.7x + 0.6 = 0$$ Multiply both sides by -1: $$x^2 + 1.7x - 0.6 = 0$$ 13. **Solve quadratic equation:** $$x = \frac{-1.7 \pm \sqrt{1.7^2 - 4 \times 1 \times (-0.6)}}{2} = \frac{-1.7 \pm \sqrt{2.89 + 2.4}}{2} = \frac{-1.7 \pm \sqrt{5.29}}{2}$$ $$\sqrt{5.29} = 2.3$$ So: $$x_1 = \frac{-1.7 + 2.3}{2} = \frac{0.6}{2} = 0.3$$ $$x_2 = \frac{-1.7 - 2.3}{2} = \frac{-4}{2} = -2$$ 14. **Interpret solutions:** Negative moles added is not possible, so discard $x = -2$. 15. **Final answer:** $$\boxed{0.3}$$ moles of CO2 must be added to reach the new equilibrium with 0.6 moles of CO2.