Opposing reactions. D. S. Martin has considered the following reversible reactio

Question

Opposing reactions. D. S. Martin has considered the following reversible reaction: Pt(NH_3)_3Cl^+ + Br^- Pt(NH_3)_3Br^+ + Cl^- The following situation can be reconstructed. At lambda = 285 nm, a solution of 1.00 times 10^-3 M Pt(NH_3)_3Cl^+ in a 5-cm cell has an absorbance of 0.450. The molar absorptivity of Pt(NH_3)_3Br^+ is 240 L mol^-1 cm^-1. The reaction was studied at 25 degree C under these concentrations: [Pt(NH_3)_3Cl^+]_0 = 1.00 times 10^-3. [Cl^-]_0 = 0.200, and [Br^-]_0 = 0.0500 M. Assuming the reaction shown defines the rate law. evaluate the forward and reverse rate constants at 25.0 degree C. (Save the results for Problem 7.9.)

Explanation / Answer

pt(NH3)3Cl+ + Br--> pt(NH3)3Br+ + Cl-

Rate of change of conc of Pt compd = Am't of cisplatin reacting (mol/L)/elapsed time (t)

The rate of reaction is proportional to [Pt(NH3)2Cl2]

This is expressed as a RATE LAW ; Rate of reaction = k [Pt(NH3)2Cl2]; k is the RATE CONSTANT

k is independent of conc. But increases with T

Rate of reaction = k [Pt(NH3)2Cl2]

=240 L mol-1 cm-1, [pt(NH3)3Cl+]0 = 1 × 10-3 M, [Cl-]0 = 0.200M, [Br-] = 0.05M

A = bc

C = A/b

C = 0.450/240 L mol-1 cm-1 = 0.001875M

Similarly calculate concentration of cis-platin at all time intervals,

C = 0.550/240 L mol-1 cm-1 = 0.00229M

C = 0.632/240 L mol-1 cm-1 = 0.00263M

C= 0.688/240 L mol-1 cm-1 = 0.00286M

C= 0.784/240 L mol-1 cm-1 = 0.00326M

C= 0.843/240 L mol-1 cm-1 = 0.00351M

To calculate K , ln [A]/[A]o = -Kt

Putting the values, A= 0.00351M; [A]o = 0.001875M; t= 1800 sec

ln 0.00351/0.00187 = K * 1800

Rate constant = K = 3.502 × 10-4

For K-1 can be calculated in terms of increase in concentration of pt(NH3)3Cl+; as pt(NH3)3Br get converted back to pt(NH3)3Cl+.

Formula will be ln [pt(NH3)3Br] / [pt(NH3)3Br]o = -K-1×t

Since here reaction going in opposite direction, at 1800 seconds, [pt(NH3)3Br]o = 0.00351M and

Final concentration at 0 seconds is , [pt(NH3)3Br]o = 0.001875M and t= 1800-600=1200 sec

Putting values in above formula ,

ln [pt(NH3)3Br] / [pt(NH3)3Br]o = -K-1×t

ln0.00351M / 0.001875M = -K-1 × 1200

K-1 = 2.512

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