1.

A sealed container was filled with 1 mol of A_(2)(g) 1 mol B_(2)(g) at 800 K and total pressure 1.00 bar. Calculate the amounts of the components in the mixture at equilibrium given that K=1 for the reaction A_(2)(g)+B_(2)(g)hArr2AB(g)

Answer»

Solution :`A_(2)(g)+B_(2)(g)hArr2AB(g)`

TOTAL no. of moles `=1-x+1-x+2X=2`
`K_(P)=((P_(AB))^(2))/((P_(A_(2)))(P_(B_(2))))=(((2x)/(2)xxP)^(2))/((((1-x))/(2)xxP)((1-x)/(2)xxP))`
`K_(P)=(4X^(2))/((1-x)^(2))`
Given that `K_(P)=1,(4x^(2))/((1-x)^(2))=1`
`rArr4x^(2)=(1-x)^(2)`
`rArr4x^(2)=1+x^(2)-2x`
`3x^(2)+2x-1=0`
`x=(-2pmsqrt(4-(4xx3xx-1)))/(2(3))`
`x=(-2pmsqrt(4+12))/(6)`
`=(-2pmsqrt(16))/(6)`
`=(2)/(6),(-2-4)/(6)`
`x=0.33,-1("not POSSIBLE")`
`:.[A_(2)]_(eq)=1-x=1-0.33=0.67`
`[B_(2)]_(eq)=1-x=1-0.33=0.67`
`[AB]_(eq)=2x=2xx0.33=0.66`.


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