Question

1. Blood chemistry (again). The chemical equation for the binding of oxygen in your blood can be written as follows: Hemoglobin(aq) + O2(aq) → Hemoglobin* O2(aq) The reaction was studied using initial rates and the following data was obtained: [Hemoglobin] [02] Initial Rate 0.0050 M 0.0100 M 2105 M/s 0.0050 M 0.0200 M 4201 M/s 0.0100 M 0.0100 M 4198 M/s a. Determine the rate law for binding of hemoglobin and oxygen b. What is the rate constant? c. What is the rate of oxygen binding to hemoglobin at physiological conditions? (Hemoglobin) = 2.0 x 10-9 Mand (O2) = 5.0 x 10-5 M. d. In HW 18, you determined that the rate constant for the binding between hemoglobin and carbon monoxide is 0.280 um 151. Is the binding between oxygen and hemoglobin a slower or faster process than the binding between carbon monoxide and hemoglobin? Justify your answer in 1-2 sentences. e. Later in class, we learned that this reaction actually exists with a double arrow () in equilibrium with the reverse reaction. If the reverse rate constant for the reaction written above is 20 s-4, calculate the equilibrium constant for this system.

Answer 1

we need to assume rate law hate = k [Hall Toy H₂ = Hemoglobin Qlos k [0.0050] (0.0100)* -0 4201 = K [0.0050)*(0.02.00] -0 dinde @ by @ apo - [0.0100] Forozoly Similarly K [p, 01607* 70.0100]* ® 4198 01 - Scanilled with CamScanner

Page No.: Rate law, - hete = k [hu] [ou b. but nalme of [Hin] & equation thn & [₂) [ ] ; in any 2105_=_k [0.00 510.01) x = elos Is421 3105 H a 4.21x10° rate = 4.21*100 ml.sh [2x103] [5x10 = 9.21X10-8 mgry do I as rate as k rate constant K should increase & decrease increases, decrease - K = 0.280 M Mis = 2.20 X10* Misal should be slower the process

FJy Equilibrim Const ke? Ky nd rate reverse rate Constant Constant = K = 4.21 XiOS MISA 20 szt Kc = 2.105 Xlo