1.

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

Answer»

<P>0.19 ATM
0.38 atm
0.6 atm
0.072 atm

Solution :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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