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A 10 moles of pcl5 are heated in a vessel of 1litre capacity At equilibrium the degree of dissociation of pcl5 is 0.6find kc?
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Introduction
To find the equilibrium constant (Kc) for the dissociation of PCl5, we start with the balanced reaction:
PCl5(g) ⇌ PCl3(g) + Cl2(g)
Initial Conditions
- Number of moles of PCl5 = 10 moles
- Volume of the vessel = 1 L
- Initial concentration of PCl5 = 10 moles/L = 10 M
Degree of Dissociation
- Given degree of dissociation (α) = 0.6
- At equilibrium, the amount of PCl5 that dissociates = α × Initial moles = 0.6 × 10 = 6 moles
Equilibrium Concentrations
- Moles of PCl5 remaining = Initial moles - Moles dissociated
- = 10 - 6 = 4 moles
- Moles of PCl3 formed = 6 moles
- Moles of Cl2 formed = 6 moles
Since the volume of the vessel is 1 L, the concentrations at equilibrium are:
- [PCl5] = 4 moles/1 L = 4 M
- [PCl3] = 6 moles/1 L = 6 M
- [Cl2] = 6 moles/1 L = 6 M
Expression for Kc
The expression for the equilibrium constant (Kc) is given by:
Kc = [PCl3][Cl2] / [PCl5]
Calculating Kc
Substituting the equilibrium concentrations into the expression:
Kc = (6)(6) / (4)
Kc = 36 / 4 = 9
Conclusion
The equilibrium constant Kc for the dissociation of PCl5 at the given conditions is:
Kc = 9
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A 10 moles of pcl5 are heated in a vessel of 1litre capacity At equilibrium the degree of dissociation of pcl5 is 0.6find kc?
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