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1mol of Cl(2) and 3 mol of PCl(5) are pl...

`1`mol of `Cl_(2)` and `3` mol of `PCl_(5)` are placed in a `100 L` vessel heated to `227^(@)C`. The equilibrium pressure is `2.05` atm. Assuming ideal behaviour, calculate the degree of dissociation for `PCl_(5)` and `K_(p)` for the reaction.
`PCl_(5)(g) hArr PCl_(3)(g)+Cl_(2)(g)`

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To solve the problem, we will follow these steps: ### Step 1: Calculate the total number of moles (N) at equilibrium using the ideal gas law. The ideal gas law is given by the equation: \[ PV = nRT \] ...
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For the reaction : PCl_(5) (g) rarrPCl_(3) (g) +Cl_(2)(g) :

For the reaction : PCl_(5) (g) rarrPCl_(3) (g) +Cl_(2)(g) :

1 mole of N_2 and 3 moles of PCl_5 are placed in a 100 litre vessels heated at 227^(@) C the equilibrium pressure is 2.05 atm Assuming ideal behaviour,Calculate degree of dissociation of PCl_5 and K_p for the reaction PCl_5 (g) hArr PCl_3 (g) + Cl_2(g)

For the reaction PCl_(5)(g) rightarrow PCl_(3)(g) + Cl_(2)(g)

Unit of equilibrium constant K_p for the reaction PCl_5(g) hArr PCl_3(g)+ Cl_2(g) is

(K_p/K_c) for the given equilibrium is [PCl_5 (g) hArrPCl_3(g) + Cl_2 (g)]

1 mole of PCl_(5) taken at 5 atm, dissociates into PCl_(3) and Cl_(2) to the extent of 50% PCl_(5)(g)hArr PCl_(3)(g)+Cl_(2)(g) Thus K_(p) is :

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For the gas-phase decomposition, PCl_(5)(g)LeftrightarrowPCl_(3)(g)+Cl_(2)(g)

PCl_(5) is 10% dissociated at 1 atm. What is % dissociation at 4 atm . PCl_(5)(g)hArr PCl_(3)(g)+Cl_(2)(g)

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