Question

The rate of the reaction between hemoglobin ($Hb$) and carbon monoxide ($CO$) was studied at ${20}^{o}C$. The following data were collected with all concentration units $\mu mol/L$

(A haemoglobin concentration of $2.21\mu mol/L$ is equal to $2.21\times {10}^{-6}mol/L$)

	$[Hb{]}_0$ $\left(\mu mol/L\right)$	$[CO{]}_0$ $\left(\mu mol/L\right)$	Initial rate $\left(\mu mol{L}^{-1}{S}^{-1}\right)$
(1)	2.21	1.00	0.619
(2)	4.42	1.00	1.24
(3)	4.42	3.00	3.71

Determine the orders of this reaction with respect to $Hb$ and $CO$ and rate constant.

• A. 1${}^{st}$ order in Hb and I${}^{st}$ order in CO and rate constant is $0.140L\mu mo{l}^{-1}{s}^{-1}$
• B. 1${}^{st}$ order in Hb and 1${}^{st}$ order in CO and the rate constant is $0.280L\mu mo{l}^{-1}{s}^{-1}$
• C. 1${}^{st}$ order in Hb, 2${}^{nd}$ order CO and the rate constant is $0.35$$L\mu mo{l}^{-1}{s}^{-1}$
• D. 2${}^{nd}$ order in Hb, 2${}^{nd}$ order in CO and the rate constat is $0.24$ $L\mu mo{l}^{-1}{s}^{-1}$

Answer 1

Answer: (B)

1${}^{st}$ order in Hb and 1${}^{st}$ order in CO and the rate constant is $0.280L\mu mo{l}^{-1}{s}^{-1}$

When the concentration of $Hb$ is doubled and the concentration of $CO$ is kept constant, the rate of the reaction is doubled. Hence, the reaction is first order in [$Hb$].

When the concentration of $CO$ is tripled and the concentration of Hb is kept constant, the rate of the reaction is tripled. Hence, the reaction is first order in order in [$CO$].

The rate law expression is $r=k\left[Hb\right]\left[CO\right]$

Substitute values in the above expression:

$0.619=k\times 2.21\times 1.00$

$k=0.280L\mu mo{l}^{-1}{s}^{-1}$.

Hence, option B is correct.