Question

Answer please

urgent.....


needed help

Answer 1

Let that dissciation of  H₂O take  place  in a  volume  of  V ltr container. Let that  a  moles  of  H₂O are  taken at  starting time and  at  equilibrium x moles of  H₂ and  x/2 moles of  O₂ are  formed as shown in figure.

                               H₂O ⇒ H₂ + 1/2 O₂

at  starting                a           0       0

at  equilibrium          a-x        x        x/2

active  mass              a-x/V     x/V   x/2v

$K_c=\frac{[H_2][O_2]^{\frac{1}{2}}}{[H_2O]}\\\;\\K_c=\frac{[x/V][x/2V]^{\frac{1}{2}}}{(a-x)/V}\\\;\\K_c=\frac{x\sqrt{x}}{(a-x)\sqrt{2V}}\\\;\\Now,\\\;\\K_p=\frac{n_{H_2}\times(n_{O_2})^{\frac{1}{2}}}{n_{H_2O}}\times\frac{P}{\sum{n}}\\\;\\\sum{n}=\text{a-x+x+(x/2)=a+\;x/2}\\\;\\K_p=\frac{Px\sqrt{x}}{(a-x)(a+\frac{x}{2})\sqrt{2}}$

Now, if  the  degree  of dissociation is  α,

α = x/a

x = aα  

$K_p=\frac{Px\sqrt{x}}{(a-x)(a+\frac{x}{2})\sqrt{2}}\\\;\\K_p=\frac{Pa\alpha\sqrt{a\alpha}}{(a-a\alpha)(a+\frac{a\alpha}{2})\sqrt{2}}K_p=\frac{Pa\alpha\sqrt{a\alpha}}{a^{2}(1-\alpha)(1+\frac{\alpha}{2})\sqrt{2}}\\\;\\K_p=\frac{2P\alpha\sqrt{a\alpha}}{a(1-\alpha)(2+\alpha)\sqrt{2}}\\\;\\K_p=\frac{\sqrt{2}P\alpha\sqrt{a\alpha}}{a(1-\alpha)(2+\alpha)}\\\;\\K_p=\frac{P\alpha\sqrt{2a\alpha}}{a(1-\alpha)(2+\alpha)}$