Solve this: Q 58 In a 0 25 L tube dissociation of 4 mol of NO is take place - Chemistry - Chemistry in Everyday Life

Question

Solve this:
Q 58 In a 0 25 L tube dissociation of 4 mol of NO is take place If
its degree of dissociation is 10 % The value of K P for reaction 2
N O ⇌ N 2 + O 2 is :-
(1) 1 18 2
(2) 1 8 2
(3) 1 16
(4) 1 32

Vartika Jain · Accepted Answer

Dear Student,

NO           ↔           12N2         +        12O2Initial              40
25                          0                         0                At equil
      40
25(1-α)            4α2×0
25             4α2×0
25       Kc = [O2]12[N2]12      [NO]It is given that, α = 0
10So, [O2]=[N2] = 4×0 102×0
25=45 [NO]=40 25(1-0 10)  =4×0 900
25=725  So, Kc =(45)12(45)12(725)=472=118Now, Kp = Kc(RT)∆nFor the given reaction, ∆n=0So, Kp=Kc =118There seems to be some problem in the options given, it should be A but instead of (18)2 it should be 18

Solve this: Q.58. In a 0.25 L tube dissociation of 4 mol of NO is take place. If its degree of dissociation is 10 %. The value of K P for reaction 2 N O ⇌ N 2 + O 2 is :- (1) 1 18 2 (2) 1 8 2 (3) 1 16 (4) 1 32

Solve this:
Q.58. In a 0.25 L tube dissociation of 4 mol of NO is take place. If its degree of dissociation is 10 %. The value of $K_{P}$ for reaction $2 N O ⇌ N_{2} + O_{2}$ is :-
(1) $\frac{1}{{(18)}^{2}}$
(2) $\frac{1}{{(8)}^{2}}$
(3) $\frac{1}{16}$
(4) $\frac{1}{32}$