Question

In a 0.25 l tube dissociation of 4 mol of no is take place if its degrer of dissociation is 10 %the value of kp for the reaction 2no gives n2 +o2

Answer 1

                    $NO\rightarrow \frac{1}{2}N_{2}+\frac{1}{2}O_{2}$

Initial     $\frac{4}{0.25}$                                              0                                    0

 

Final       $\frac{4\times (1-\alpha )}{0.25 }        \frac{4\times \alpha }{2\times 0.25 } \frac{4\times \alpha }{2\times 0.25 }$

 $Kc=\frac{O_{2}^{1/2}\times N_{2}^{1/2}}{NO}$

It is given that it dissociate only 10 %, so $\alpha$ here is 0.01

So, [O]=[N]=$\frac{4\times (0.10 )}{2\times 0.25 }=\frac{4}{5}$

 

[NO]=$\frac{4\times (1-0.10 )}{ 0.25 }=\frac{4\times (0.90 )}{0.25}=\frac{72}{5}$

Putting these values in Kc expression:

$Kc=Kc=\frac{(\frac{4}{5})^{1/2}\times (\frac{4}{5})^{1/2}}{\frac{72}{5}}=\frac{1}{18}$  

Kp is related to kc by the following equation:

 $Kp=Kc \times (RT)^{\Delta n}$  

Here, ${\Delta n}$ is 0

So  Kp=Kc=$\frac{1}{18}$  

so Kp=Kc=$\frac{1}{18}$  

Answer 2

hope it will help you