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Grade: 11

                        

At 90°C , the following equilibrium is established : H2 (g) + S(s) = H2 S(g) Kp = 6.8 × 10 ^2 . If 0.2 mol of hydrogen and 1.0 mol of sulphur are heated to 90°C in a 1.0 litre vessel, what will be the partial pressure of H2 S at equilibrium?

7 years ago

Answers : (3)

Arpit Jaiswal
31 Points
							

given,

H2 (g) +    S(s)      = H2 S(g)     Kp  = 6.8 × 10 ^2 

int. moles     0.2          1.0           ----

final moles   (0.2-x)    (1.0-x)        x

total moles are 1.2-x

therefore, from Kp equation we have-

6.8 x 10^2  =                   [x/(1.2-x) * 1]              

[(0.2-x)/(1.2-x)*1][(1-x)/(1.2-x)*1]

solving the above quadratic, we get- x

now partial pressure of H2 S is  (x/1.2-x)*1

which is the required answer

hope this helps

AJ

 

 

7 years ago
ankitesh gupta
63 Points
							

 

 

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7 years ago
ankit singh
askIITians Faculty
596 Points
							

H2(g)+S(s)H2S(g)
0.2        2              0
0.2-x      2-x           x 

Δn=0,Kp=Kc 

Kc=[H2][H2S]=(0.2x)x=6.8×102 


On solving this, we get
 
=>x=1.27×102
 
PH2S=[x]RT=1.27×102×0.0821×(273+90)=0.38atm
2 months ago
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