Chemistry, asked by adilarif1998p45pce, 1 year ago

the degree of dissociation of pcl5 at 16.8 bar and 127°C is 0.4 find the Kp for the reaction

Answers

Answered by DxddyExplicit
0
The Kp is equal to 139.4 C
Answered by BarrettArcher
1

Answer : The equilibrium constant for the reaction is, 3.2

Solution :  Given,

Total pressure = 16.8 bar

Degree of dissociation, \alpha = 0.4

The given equilibrium reaction is,

                            PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)

Initially                   1                  0                  0

At equilibrium    (1-\alpha)            \alpha                 \alpha

\text{ Total number of moles}=1-\alpha+2\alpha=1+\alpha

Now we have to calculate the partial pressure of PCl_5, PCl_3 and Cl_2

\text{ Partial pressure of }PCl_5=\frac{\text{Moles of }PCl_5}{\text{Total number of moles}}\times P=\frac{(1-\alpha)}{(1+\alpha)}\times P

\text{ Partial pressure of }PCl_3=\frac{\text{Moles of }PCl_3}{\text{Total number of moles}}\times P=\frac{(\alpha)}{(1+\alpha)}\times P

\text{ Partial pressure of }Cl_2=\frac{\text{Moles of }Cl_2}{\text{Total number of moles}}\times P=\frac{(\alpha)}{(1+\alpha)}\times P

The expression of K_p will be,

K_p=\frac{(p_{PCl_3})(p_{Cl_2})}{p_{PCl_5}}

K_p=\frac{(\frac{(1-\alpha)}{(1+\alpha)}\times P)(\frac{(\alpha)}{(1+\alpha)}\times P)}{\frac{(\alpha)}{(1+\alpha)}\times P}

K_P=\frac{\alpha^2P}{(1-\alpha^2)}

Now put all the values in this expression, we get

K_P=\frac{(0.4)^2\times 16.8}{[1-(0.4)^2]}=3.2

Therefore, the equilibrium constant for the reaction is, 3.2

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