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Consider the reaction X(g) hArr Y(g)+Z...

Consider the reaction
`X(g) hArr Y(g)+Z(g)`
When the system is at equilibrium at `100^(@)`, the concentrations are found to be `[X]=0.2 M, [Y]=[Z]=0.4 M`
a. If the pressure of the container is suddenly halved at `100^(@)C`, find equilibrium concentration.
b. If the pressure of the container is suddenly doubled at `100^(@)C`, find the equilibrium concentration

Text Solution

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`{:(X,hArr,Y,+,Z,),(0.2,,0.4,,0.4,"At equilibrium"):}`
`K_(c )=(0.4xx0.4)/0.2=0.8`
a. When pressure is halved, `(i.e. P rarr P/2)`, concentration is halved.
`:. [X]=0.1 M, [Y]=[Z]=0.2 M`
`Q_(c )=(0.2xx0.2)/(0.1)=0.4`
`:. Q_(c ) lt K_(c )` (Reaction goes forward)
`K_(c )=0.8=((0.2+x)(0.2+x))/((0.1-x))`
solving for `x`:
`x=(-0.4+sqrt(0.16+4xx0.04))/2=0.08`
`:.` Equilibrium concentration:
`[X]=0.1-x=0.1-0.08=0.02 M`
`[Y]=[Z]=0.2+x=0.2+0.08=0.28 M`
b. When pressure is doubled (i.e., `P rarr 2P)`,
concentration is doubled.
`:. [X]=0.4 M [Y]=[Z]=0.8 M`
`Q_(c)=(0.8xx0.8)/0.4=1.6`
`:. Q_(c) gt K_(c)` (Reaction goes backward)
`K_(c)=0.8=((0.8-x)(0.8-x))/((0.4+x))`
Solving for x:
`x^(2)-0.4x+0.32=0`
`x=(-(-0.4)+sqrt(0.16-1.28))/(2)~~0.73`
`:.` Equilibrium concentration:
`[X]=0.4+x=0.4+0.73=1.13 M`
`[Y]=[Z]=0.8-x=0.8-0.73=0.07 M`
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