The rate of the haemoglobin (Hb)- carbon monoxide reaction ,
`4Hb+3CO to Hb_(4)(CO)_(3)`
has been studied at `20^(@)C` .Concentrations are expressed in `mu ` mole/L .
`{:([Hb](mu"mole/L"),[CO](mu"mole/L"),"Rate of disappearance of Hb"(mu"mole/L/s")),(3.36,1.00,0.941),(6.72,1.00,1.88),(6.72,3.00,5.64):}`
Calculate the rate of the reaction at the instant when
`[Hb] =1.50 and [CO] =0.60 mu "mole /L"`

Explanation

To solve the problem, we need to calculate the rate of the reaction at the instant when the concentrations of Hb and CO are given as [Hb] = 1.50 µmole/L and [CO] = 0.60 µmole/L. We will follow these steps:

### Step 1: Write the rate expression for the reaction
The reaction is given as:
\[ 4 \text{Hb} + 3 \text{CO} \rightarrow \text{Hb}_4(\text{CO})_3 \]

The rate of the reaction can be expressed in terms of the disappearance of Hb:
\[ \text{Rate} = -\frac{1}{4} \frac{d[\text{Hb}]}{dt} \]

### Step 2: Determine the rate law
The rate law can be expressed as:
\[ \text{Rate} = k [\text{Hb}]^m [\text{CO}]^n \]
where \( k \) is the rate constant, and \( m \) and \( n \) are the orders of the reaction with respect to Hb and CO, respectively.

### Step 3: Use the provided data to find \( m \) and \( n \)
We will use the provided data points to find the values of \( m \) and \( n \).

1. From the first data point:
\[ [\text{Hb}] = 3.36 \, \mu\text{mole/L}, \quad [\text{CO}] = 1.00 \, \mu\text{mole/L}, \quad \text{Rate} = 0.941 \, \mu\text{mole/L/s} \]
\[ \frac{0.941}{4} = k (3.36)^m (1.00)^n \]

2. From the second data point:
\[ [\text{Hb}] = 6.72 \, \mu\text{mole/L}, \quad [\text{CO}] = 1.00 \, \mu\text{mole/L}, \quad \text{Rate} = 1.88 \, \mu\text{mole/L/s} \]
\[ \frac{1.88}{4} = k (6.72)^m (1.00)^n \]

3. From the third data point:
\[ [\text{Hb}] = 6.72 \, \mu\text{mole/L}, \quad [\text{CO}] = 3.00 \, \mu\text{mole/L}, \quad \text{Rate} = 5.64 \, \mu\text{mole/L/s} \]
\[ \frac{5.64}{4} = k (6.72)^m (3.00)^n \]

### Step 4: Solve for \( m \)
Using the first two equations, we can eliminate \( k \) and solve for \( m \):
\[ \frac{0.941/4}{1.88/4} = \frac{(3.36)^m}{(6.72)^m} \]
This simplifies to:
\[ \frac{0.941}{1.88} = \left(\frac{3.36}{6.72}\right)^m \]
Calculating the left side gives approximately 0.5. The right side simplifies to \( (0.5)^m \). Thus:
\[ 0.5 = (0.5)^m \implies m = 1 \]

### Step 5: Solve for \( n \)
Using the second and third equations:
\[ \frac{1.88/4}{5.64/4} = \frac{(6.72)^m (1.00)^n}{(6.72)^m (3.00)^n} \]
This simplifies to:
\[ \frac{1.88}{5.64} = \left(\frac{1}{3}\right)^n \]
Calculating the left side gives approximately 0.333. Thus:
\[ 0.333 = \left(\frac{1}{3}\right)^n \implies n = 1 \]

### Step 6: Calculate the rate constant \( k \)
Using any of the data points, we can find \( k \). Using the first data point:
\[ \frac{0.941}{4} = k (3.36)^1 (1.00)^1 \]
Calculating gives:
\[ k = \frac{0.941/4}{3.36} \approx 0.070 \]

### Step 7: Calculate the rate at the given concentrations
Now we can find the rate at [Hb] = 1.50 µmole/L and [CO] = 0.60 µmole/L:
\[ \text{Rate} = k [\text{Hb}]^1 [\text{CO}]^1 = 0.070 \times 1.50 \times 0.60 \]
Calculating this gives:
\[ \text{Rate} \approx 0.063 \, \mu\text{mole/L/s} \]

### Final Answer:
The rate of the reaction at the instant when \([Hb] = 1.50 \, \mu\text{mole/L}\) and \([CO] = 0.60 \, \mu\text{mole/L}\) is approximately \( 0.063 \, \mu\text{mole/L/s} \).