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

If 0.2 mol of `H_(2(g))` and 2.0 mol of `S_((s))` are mixed in a `1dm^(3)` vessel at `90^(@)C`, the partial pressure of `H_(2)S_((g))` formed according to the reaction
`H_(2(g))+S_((s))iffH_(2)S,K_(p)=6.8xx10^(-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_(2)` have reacted, then at eqm.,
`[H_(2)]=(0.2-x),[H_(2)S]=x`
`p_(H_(2))=(0.2-x)/(0.2-x+x)=(x)/(0.2)xxP`
`K_(p)=(p_(H_(2)S))/(p_(H_(2)))i.e.,6.8xx10^(-2)=(x)/((0.2-x))`
or, `0.068(0.2-x)=xorx=0.0127mol`
Pressure of 0.0127 mol of `H_(2)S` at 363 K in 1 L vessel,
`P=(nRT)/(V)=(0.0127xx0.0821xx363)/(1)=0.38atm`
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