男士手表国际品牌排行榜前十名 The rote of the reaction between hemoglobin (Hb) and carbon monoxide (CO) was studied at 20°C. The following data were collected with all concentration units in μ mol/L. (A hemoglobin concentration of 2.21 μ mol/L is equal to 2.21 × 10 −6 mol/L.) [Hb] 0 ( μ mol/L) [CO] 0 ( μ mol/L) Initial Rate ( μ mol/L · s) 2.21 1.00 0.619 4.42 1.00 1.24 4.42 3.00 3.71 a. Determine the orders of this reaction with respect to Hb and CO. b. Determine the rate law. c. Calculate the value of the rate constant. d. What would be the initial rate for an experiment with [Hb] 0 = 3.36 μ mol/L and [CO] 0 = 2.40 μ mol/L?

The rate law for the given reaction is calculated by the expression,

Rate=k[Hb]m[CO]n

Where,

[Hb] and [CO] are the concentrations of Hb and CO . k is the rate constant.

The values of m and n are calculated by the comparison of the different rate values from the given table.

The value of m is calculated using the first and second result. Substitute the values of the concentration of [Hb] and [CO] for the first two experiments in the above expression.

Rate 1=k[2.21]m[1.00]n

Rate 2=k[4.42]m[1.00]n

According to the given rate values in the table,

Rate 2Rate 1=1.24 μmol/L⋅s0.619 μmol/L⋅s

Therefore,

k[4.42]m[1.00]nk[2.21]m[1.00]n=1.24 μmol/L⋅s0.619 μmol/L⋅s(2)m=2m=1_

The value of n is calculated using the second and third result. Substitute the values of the concentration of [Hb] and [CO] for the second and third experiments in the above expression.

Rate 2=k[4.42]m[1.00]n

Rate 3=k[4.42]m[3.00]n

According to the given rate values in the table,

Rate 3Rate 2=3.71 μmol/L⋅s1.24 μmol/L⋅s

Therefore,

k[4.42]m[3.00]nk[4.42]m[1.00]n=3.71 μmol/L⋅s1.24 μmol/L⋅s(3)n=3n=1_

Substitute the values of m and n in the rate law expression.

Rate=k[Hb]1[CO]1