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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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