CHEM 104 Homework 3 4. (20 points): Chapter 16.3 Hemoglobin, the main protein th
ID: 577563 • Letter: C
Question
CHEM 104 Homework 3 4. (20 points): Chapter 16.3 Hemoglobin, the main protein that carries oxygen in the blood, can also bind carbon monoxide molecules, which is the main mechanism of carbon monoxide poisoning if a person inhales too much CO gas. The following data were collected in a study of the rate of the reaction between hemoglobin (Hb) and carbon monoxide (CO) at 20°C All concentration units are given in units ot umoVL (1 ol /Le 106 mol/L). 2.21 4.42 4.42 1.00 1.00 3.00 0.619 1.24 3.71 a. Determine the orders of this reaction with respect to Hb and CO. b. Determine the rate law and overall order of the reaction. c. Calculate the value of the rate constant. Be sure to include correct units! d. What would be the initial rate of reaction for an experiment with the following conditions: [Hblo-2.50 moVL and [colo= 1.75 mol/L?Explanation / Answer
Consider a rate law for the given reaction
Rate = K [Hb]ao[CO]bo -------(1)
K = rate constant, a is order of reaction with respect to Hb
b is order of reaction with respect to CO
From run 1:- We can write with the help of (1)
0.619 * 10^-6 mol/L.s = K(2.21 * 10^-6 mol/L)^a(1.00 * 10^-6mol/L)^b. ------(2)
From run 2:- we can write with the help of (1)
1.24 * 10^-6 mol/L.s = K(4.42*10^-6 mol/L)^a(1.00*10^-6 mol/L)^b -----(3)
From run 3:- we can write with the help of (1)
3.71*10^-6 mol/L.s = K(4.42*10^-6 mol/L)^a(3.00*10^-6 mol/L)^b ------(4)
Divide (3) by (2)
(1.24*10^-6 mol/L.s)/(0.619*10^-6 mol/L.s) = {K(4.42*10^-6 mol/L)^a(1.00*10^-6 mol/L)^b / K(2.21*10^-6 mol/L)^a(1.00*10^-6 mol/L)^b}
(2.00)^1 = (2.00)^a
a = 1
Divide (4) by (3)
3.71*10^-6 mol/L.s / 1.24 * 10^-6 mol/L.s = {K(4.42*10^-6 mol/L)^a(3.00*10^-6 mol/L)^b / K(4.42*10^-6 mol/L)^a(1.00*10^-6 mol/L)^b}
(3.00)^1 = (3.00)^b
b = 1
(a) order of reaction with respect to Hb is 1(first order).
Order of reaction with respect to CO is 1(first order)
(b) Rate law from (1)
Rate = K[Hb]o[CO]o -----(5)
Overall order of reaction = 1+1 = 2(second order)
(c) put all the values in (2) to calculate rate constant (K)
0.619*10^-6 mol/L.s = K(2.21*10^-6 mol/L)(1.00*10^-6 mol/L)
0.619*10^-6 mol/L.s = K(2.21*10^-12 mol^2 L^-2)
K = 0.619*10^-6 mol L^-1 s^-1 / 2.21*10^-12 mol^2 L^-2
K = 0.28*10^6 mol^-1 L s^-1
K = 2.8*10^5 mol^-1 L s^-1
(d) Put all values in (5)
Rate = 2.8*10^5 mol^-1 L s^-1 (2.5*10^-6 mol/L)(1.75*10^-6 mol/L)
Rate = 12.25 * 10^-7 mol/L.s = 1.225 micro mol /L.s