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# he equilibrium constant KP for the reaction:$PC{l_5} \rightleftharpoons PC{l_3} + C{l_2}$ Is $1.6$ at $200^\circ C$.The pressure at which $PC{l_5}$ will be $50\%$ dissociated at $200^\circ C$ is.

Last updated date: 01st Mar 2024
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Hint: We know that the equilibrium constant of a synthetic response is the worth of its response remainder at substance balance; a state drew nearer by a powerful compound system after adequate time has slipped by at which its piece has no quantifiable inclination towards additional change. For a given arrangement of response conditions, the balance consistent is autonomous of the underlying scientific convergences of the reactant and item species in the blend. Accordingly, given the underlying organization of a system, known balanced steady qualities can be utilized to decide the piece of the system at equilibrium. Be that as it may, response boundaries like temperature, dissolvable, and ionic strength may all impact the worth of the balance of steady concentration.

The balanced given equation is,
$PC{l_5} \rightleftharpoons PC{l_3} + C{l_2}$
Let us assume the initial concentration of phosphorus pentachloride as one mole.
It is given $\alpha = 0.5$
The ICE table is,
 $PC{l_5}$ $PC{l_3}$ $C{l_2}$ $1$ $0$ $0$ $1 - 0.5$ $0.5$ $0.5$

The number of moles at equilibrium is calculated as,
Total number of moles$= 0.5 + 0.5 + 0.5 = \dfrac{3}{2}$
The equilibrium constant Kp is calculated as,
${k_P} = \dfrac{{\dfrac{1}{2}\left( {\dfrac{1}{2}} \right)}}{{\dfrac{1}{2}}} \times \dfrac{P}{{\dfrac{3}{2}}}$
$1.6 = \dfrac{1}{4} \times 2 \times \dfrac{{2P}}{3}$
On multiplication we get,
$P = 1.6(3)$
On simplification we get,
$\Rightarrow P = 4.8atm$
The pressure is calculated as $4.8\,atm$.

Note:
Now we can discuss about the attributes of equilibrium constant as,
It is response explicit and at a consistent temperature, it is fixed.
An impetus changes the pace of forward and in reverse responses similarly not to influence the worth of the balance consistently.
Changes in focus, pressure, temperature, idle gases may influence the balance, preferring either forward or in reverse response however not the equilibrium constant.
Is identified with the standard free energy as,$\Delta {G_0} = - {\text{ }}RTln{\text{ }}{K_{equ}}$.
For a similar reversible response, ${K_{equ}}$ have various qualities at various temperatures.
The balance consistent with the opposite equilibrium is the equal of the first balance for example ${K_{rev}} = 1/{K_{equ}}$.
In the event that the balance response stoichiometry is changed, the force of the equilibrium steady additionally gets changed by a similar amount.
Concurrent balance responses, having a typical item. The equilibrium steady of the responses doesn't change. Because of the greater convergence of the regular item, the item fixations will be diminished.