Chapter 15: Problem 87
Phosgene, \(\mathrm{COCl}_{2}\), is a toxic gas used in the manufacture of urethane plastics. The gas dissociates at high temperature. $$ \mathrm{COCl}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{Cl}_{2}(g) $$ At \(400^{\circ} \mathrm{C}\), the equilibrium constant \(K_{c}\) is \(8.05 \times 10^{-4}\). Find the percentage of phosgene that dissociates at this temperature when \(1.00\) mol of phosgene is placed in a \(25.0\) - \(\mathrm{L}\) vessel.
Short Answer
Step by step solution
Identify Initial Conditions
Set up the ICE Table
Write the Expression for Equilibrium Constant
Solve for x
Calculate the Percentage Dissociation
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Phosgene Dissociation
Equilibrium Constant
- A smaller \( K_c \) value indicates the reaction does not go to completion and remains mostly as reactants.
- It is a measure that assists chemists in predicting the extent of dissociation under given conditions.
- Understanding \( K_c \) is vital for controlling the conditions to enhance or inhibit the dissociation based on the desired chemical process outcome.
ICE Table
- Initial: \([\text{COCl}_2] = 0.040\,\text{mol/L}, \,[\text{CO}] = 0, \,[\text{Cl}_2] = 0\)
- Change: \([\text{COCl}_2] = -x, \,[\text{CO}] = +x, \,[\text{Cl}_2] = +x\)
- Equilibrium: \([\text{COCl}_2] = 0.040 - x, \,[\text{CO}] = x, \,[\text{Cl}_2] = x\)
Percentage Dissociation
- Change in concentration, \( x = 0.00567 \).
- Initial concentration = \( 0.040 \text{ mol/L} \).
- Percentage dissociation is \( \frac{0.00567}{0.040} \times 100\% \approx 14.18\% \).