To measure the equilibrium constant K for the reaction A_2 + B_2 doubleheadarrow

Question

To measure the equilibrium constant K for the reaction A_2 + B_2 doubleheadarrow 2 AB, a student reacts 20 mL of 0.035 M A_2 with 35 mL of 0.075 M B_2. A portion of the mixture produced an absorbance value of 0.357 for AB when analyzed with visible absorption spectroscopy. If the student's Beer's Law plot has slope of 79.49 M^-1, determine a. The initial molar concentration of A_2 b. the initial molar concentration of B_2. c. The equilibrium molar concentration of AB. d. the equilibrium molar concentration of A_2. e. the equilibrium molar concentration of Br_2. f. the equilibrium constant K_c for the reaction.

Explanation / Answer

The plot of Absorbance Vs concentration is A= ebC. So the plot of A vs C gives the value eb. ( e is molar extinction coefficient, b is path length, C is concentration)

Since eb is the slope of absorbance vs concentration plot, C= A/eb= 0.357/79.49 M is the concentration of AB. C= 0.0045

Moles of A2 initially= Molarity* Volume (L)= 0.035*20/1000 =0.0007, moles of B2 initially= 0.075*35/1000 =0.002625

Volume after mixing = 20+35= 55ml= 55/1000 L=0.055L

Initial concentrations : A2= 0.0007/0.055=0.0127M, B2= 0.002625/0.055=0.0477

As per the reaction A2+B2<----->2AB, 1 mole of A2 reacts with 1 mole of B2 to gives 2moles of AB.

since A2:B2 is   0.007 :0.002625, excess is B2. Limiting reactant is A2.

There are 0.0045 M of AB at equilibrium which consumes 0.0045/2 M of A2.

A2 consumed to reach equilibrium = 0.00225

At equilibrium [A2] =0.0127-0.0025= 0.0102, [B2] =0.0477-0.0025=0.0452 M, [AB]= 0.0045

K= equilibrium constant =[AB]2/ [A2] [B2] = 0.0045*0.0045/ (0.0452*0.0102)= 0.0439

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