Question

1 )for the equilibrium AB---->A+B kp is equal to 4 times of the total pressure . the no. of moles of A formed
1) root 5/2 2) 2 3) root 5 4) 2/root 5

Answer 1

Answer : The correct option is, (4) $\frac{2}{\sqrt{5}}$

Solution :  Given,

The given equilibrium reaction is,

                           $AB(g)\rightleftharpoons A(g)+B(g)$

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 $AB$, $A$ and $B$

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

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

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

The expression of $K_p$ will be,

$K_p=\frac{(p_{A})(p_{B})}{p_{AB}}$

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

$K_p=\frac{\alpha^2P}{(1-\alpha^2)}$      ..............(1)

As we are given that,

$K_p=4\times P$         .................(2)

Now equating equation 1 and 2, we get

$4=\frac{\alpha^2}{(1-\alpha^2)}$

$\alpha=0.82=\frac{2}{\sqrt{5}}$

Moles of A = $\alpha=0.82=\frac{2}{\sqrt{5}}$

Therefore, the correct option is, (4) $\frac{2}{\sqrt{5}}$

Answer 2

Answer:

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