Question

In the reaction A gives B + 2C equilibrium is established by heating 4 moles of A in a closed container at 2 atm pressure. If degree of dissociation of A is 0.25 then Kp is?

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Answer 1

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Answer 2

Answer : The value of Kp is, $1.64\times 10^{-3}$

Solution :  Given,

Degree of dissociation = $\alpha$ = 0.25

Total pressure = P = 2 atm

The given equilibrium reaction is,

                           $A(g)\rightleftharpoons B(g)+2C(g)$

Initially                  4         0           0

At equilibrium    $(4-\alpha)$     $\alpha$          $2\alpha$

$\text{ Total number of moles}=4-\alpha+3\alpha=4+2\alpha$

Now we have to calculate the partial pressure of $A$, $B$ and $C$

$\text{ Partial pressure of }A=\frac{\text{Moles of }A}{\text{Total number of moles}}\times P=\frac{(4-\alpha)}{(4+2\alpha)}\times P$

$\text{ Partial pressure of }B=\frac{\text{Moles of }B}{\text{Total number of moles}}\times P=\frac{(\alpha)}{(4+2\alpha)}\times P$

$\text{ Partial pressure of }C=\frac{\text{Moles of }C}{\text{Total number of moles}}\times P=\frac{(2\alpha)}{(4+2\alpha)}\times P$

The expression of $K_p$ will be,

$K_p=\frac{(p_{B})(p_{C})^2}{p_{A}}$

$K_p=\frac{(\frac{(\alpha)}{(4+2\alpha)}\times P)(\frac{(2\alpha)}{(4+2\alpha)^2}\times P)}{\frac{(4-\alpha)}{(4+2\alpha)}\times P}$

Now put all the given values in this formula, we get

$K_p=1.64\times 10^{-3}$

Therefore, the value of Kp is, $1.64\times 10^{-3}$