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If 0.2 mol of H(2(g)) and 2.0 mol of S(s...

If 0.2 mol of `H_(2(g))` and 2.0 mol of `S_(s)` are mixed in a 1 dm`""^3" vessel at 90"^@C` the partial pressure of `H_(2)S_(g)` formed according to the reaction, `H_(2(g)) +S_(s) Leftrightarrow, (K_(p)=6.8 xx 10^(-2))` would be

A

0.19 atm

B

0.38 atm

C

0.6 atm

D

0.072 atm

Text Solution

Verified by Experts

The correct Answer is:
B
Suppose x moles of `H_2S` have formed, then at equilibrium `[H_(2)]=(0.2-x), [H_(2)S]=x`
`p_(H_(2))=(0.2-x)/(0.2-x+x)=(0.2-x)/(0.2) xx P`
`p_(H_(2)S)=(x)/(0.2-x+x) =(x)/(0.2) xxP`
`K_(p)=(p_(H_(2))S)/(p_(H_(2))" i.e, "6.8 xx 10^(-2)) =(x)/(0.2-x)`
or 0.068 (0.2-x)=x or x=0.0127 mol
Pressure of 0.0127 mol of `H_(2)S` at 363K in 1L vessel
`P=(nRT)/V=(0.0127 xx 0.0821 xx 363)/(1)="0.38 atm"`
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