Question

D-d 7. The equation is not correctly matched for (where D = vapour density of 'A' (n-1)d before dissociation, d = vapour density of equilibrium mixture) (A) A(9) = (C) A(9) ()+(6)0(0) B(6)-@ca) (B) A(n) = (O) A() = B(9)+() C(O) B(g) + C(q)

pls do both question, if you do only one question will dislike.....

Answer 1

Answer 1
Firal moles 1-a Ang 7 (a) For the reaction - - Alg) = (2)99)+ () cg). Initial moles 1 0 1-a na na Total. no. of moles after dissociation = 1-a+40 + x = 1 +«(3n+20= 6) = 1 +(50% na na occuped mole = v litre Let volume . Dal And after dissociation volume occupied = { 1 + $n-6 - DE 1+(5n+6) . f7c5n_tajr (nb)x=ds -> 2=612 ) > a = 6 (D-d) - Ent d(50-6) (b). For the reaction Fritial mole mod (3)04) + (3)(4) + D d Intd Initial moles Final moles 1 1-0

in = 1 + (n-1). Total no. of moles after dissociation La + 2a + 2nd = 1-atina Dat and da I [1+(n-1) av = 1+0:43) = → a= D-1 nt For the reaction D-d din-1 1. Alg) - ( 2 ) Blg) + ( ) ( 19) Clearly after dissociation, no. of moles = 1-a ena na = 7+ (3n-4) 4 I So, clearly x 7 . D-d dint) For the reaction _ Aly) ) BG) + C6g) Clearly after dissociation, no. of moles sl-xt not a = 1 tha so, clearly at od So, the equation a = 2 equation a = D-d (A), (C) and (0) o is not correctly matched for d(n-1)

Answer 2

Ans (35) 2 24 N₂ + 3H₂ Nis 1 N₂ + 3H Initial 1 0 0 Final 1-4 & 3x K= IN, PH]/ - NH] → 1.49x105 = 2.13 - (x)/2. (3x)/2 - - X (1.49x105) - (1.49x105) x = 3x² solving we get = 0.26% = 0.25% So, the answer is (B)