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# 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?

8 years ago

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

hope this helps

AJ

8 years ago

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## H2​(g)+S(s)⇔H2​S(g)0.2        2              00.2-x      2-x           x ∵Δn=0,Kp​=Kc​Kc​=[H2​][H2​S]​=(0.2−x)x​=6.8×10−2

On solving this, we get

=>x=1.27×102

PH2S=[x]RT=1.27×102×0.0821×(273+90)=0.38atm