The following problem simulates the type of calculation you\'ll do to analyze th

Question

The following problem simulates the type of calculation you'll do to analyze the data from your experiment on Tuesday. The following reaction occurs: A + B rightarrow AB A and B are both colorless but AB is red. The following equation is obtained from a calibration curve of AB in a spectrophotometer (absorbance vs wavelength). y = 4.5lx + 0.00134 You mix 10 mL of 0.01 M A and 10 mL of 0.01 M B and allow the reaction time to complete, then measure the absorbance at the lambda_max, of AB. You find that the absorbance of the solution is 0.015. Write the equilibrium expression in terms of [A], [B], and [AB]. Calculate the equilibrium constant from this single trial.

Explanation / Answer

Part 1. Trendline equation of the graph is “Y = 4.51 X + 0.00134” in form of “Y = mX + C”

In the graph, Y-axis indicates absorbance and X-axis depicts concentration. That is, according to the trendline (linear regression) equation Y = 4.51 X + 0.00134 obtained from the graph, 1 absorbance unit (1 Y = Y) is equal to 4.51 units on X-axis (concentration) plus 0.00134.

Since AB is the colored product, it only absorbs light. The colorless reactants A and B do not absorb light. So, the absorbance of the solution is solely due to presence of AB.

Given,

Y = absorbance of AB = 0.015

Now, Putting the value of Y in trendline equation

            0.015 = 4.51 X + 0.00134

            Or, 4.51 X = 0.015 - 0.00134 = 0.01366

            Or, X = 0.01366 / 4.51 = 0.00302

Therefore, an absorbance of 0.015 is equivalent to 0.00302 unit concentration.

Unit of concertation = M

Thus, concentration of AB product = 0.003 M

Part 2: Total volume of reaction mixture = 10.0 mL (of A) + 10.0 mL (of B) = 20.0 mL

Doubling the volume, when both A and B are mixed in equal volumes, halves the concertation of A and B in the reaction mixture.

So, concentration of A in reaction mixture = (0.01 / 2) M = 0.005 M

Similarly, concentration of B in reaction mixture = (0.01 / 2) M = 0.005 M

Stoichiometry: 1 mol A reacts with 1 mol B to produce 1 mol AB.

Since there is formation of 0.003 M product (AB), the concertation of A and B reduces by same extent.

So,

            [A] remaining at equilibrium (after experiment) = Initial [A] – [AB] produced

                                                                                    = 0.005 M- 0.003 M = 0.002 M

            [B] remaining at equilibrium (after experiment) = Initial [B] – [AB] produced

                                                                                    = 0.005 M- 0.003 M = 0.002 M

Thus, at equilibrium (after experiment), we have –

            [A]= 0.002 M              - reactant

            [B] = 0.002 M             - reactant

            [AB] = 0.003 M          - product.

Equilibrium constant expression = [AB] / ([A] [B])              ; all [conc] at equilibrium

Now,

Equilibrium constant, K = (0.003) / (0.002 x 0.002) = 750

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