JEE Advanced Model Paper 7 with solutions for online practice

PNH4 = PHCN = 0.149 atm ⇒ Kp = 0.149 × 0.149 = 0.022 atm
In second case : NH4CN → NH3 + HCN
Initially : (0.25) , 0
At equation : (0.25 + x) , x
Kp = (0.25 + x) x = 0.022
or x2 + 0.25x - 0.022 = 0
∴ x =

−0.25+√(0.25)2−4(1)(0.022)

2(1)

=

−0.25+√0.0625−0.088

2

=

−0.25+√0.1505

2

=

−0.25+0.388

2

= 0.07
∴ [HCN] = 7 × 10−2
Hence y = 7