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`        solid NH4I on rapid heating in a closed vessel at357 degree celcius develops a const.pessue of 275 mm hg owing to partial decompositn of NH4I into NH3 and HI but pressure gradually increases further [when excess solid residue remains in vessel] owing to decompositn of HI.calculate final pressure developed at equlibrium.NH4I(s)forms NH3(g) + HI(g)2HI(g) forms H2(g)+I2(g)  Kc=0.065 at 357 degree celcius.`
4 years ago

Ashutosh Sharma
181 Points
```							NH4I(g) NH3(g) + HI(g) ------(1)using Kp and Kc relation Kp can be found out or for HI division Kp is 0.015.for this eqn 2p=275 –>p=137.5So kpfor eq. (1)Kp= P[NH3]XP[HI]= 137.5 X 137.5 ---------(2)For   2HI H2+ I21      00(1 - x)x/2 x/22HI H2+ I2 Kp= (0.015)2x = 0.2Again since HI decomposes reaction proceeds in forward directioal.But a = (P + A) X 0.2also kpfor final eqn (3)= 0.8(P + A)2But Kp= 137.5 X 137.50.8(P + A)2= 137.5 X 137.5P + A = 153.73A = 16.22a = (p + A) X 0.2 = 153.73 X 0.2 = 30.75Total pressure at eqn =P[NH3]+P[HI]+P[H2]+P[I]=(P+A)+(P+A-a)+A/2+A/2=2p+2A=2(p+A)= 2x153.73=307.46 mmHg
```
4 years ago
Akshita Goyal
17 Points
```							can u pls specify  what is A and a and how did the equatn nh4i forms nh3 (P+A)+hI(P+A-a) come?
```
4 years ago
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