The following equilibrium was studied spectrophotometrically: A+B <--> C where C

Question

The following equilibrium was studied spectrophotometrically:

A+B <--> C

where C absorbs light at 455nm. one reaction mixture was formed by mixing 25.0 mL of .10 M of A with 25mL of .10 M B and allowed to equilibriate. the percent transmittance of the solution was found to be 28. Based on the calibration curve of Y=26.1x + .081 (where absorbance is on y axis and concentration of product C is on the x axis) determine the equilibrium concentration of C, then using an ICE table format determine the value of K.

Explanation / Answer

given percent transmittance is 28 .. so we have absorbance = -log(28/100) absorbance= .2366 using Y=26.1 X + .081 substituting Y=.2366 we get .006 M of C ... using ICE table format we have initially no C .. SO all c existing during equilibrium is due to reaction of A and B .. initial conc. of A =.1 M initial conc. of B =.1 M equilibrium concentration of C= .006 M so we have equilibrium concentration of A= .1-.006 =.094 M and equilibrium concentration of B = .1 - .006 =.094 M we have K=[C]/ ([A]*[B]) USING CALCULATED VALUES WE GET K= .006 /(.094 *.094) => K= .68

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