1. Consider the reaction S -> P. At pH = 6, the reaction is zero order. At pH = 3, the reaction is first order. In the figure below the concentration product [P] is given as a func... 1. Consider the reaction S -> P. At pH = 6, the reaction is zero order. At pH = 3, the reaction is first order. In the figure below the concentration product [P] is given as a function of time. Sketch in this figure the concentration of the substrate S as a function of time. 2. Again, consider the reaction S -> P. At the beginning of an experiment you add 100 µL substrate with a concentration of 0.55 mM to a reaction mixture with a total volume of 2.3 mL. A reaction starts. After 100 s you measure the absorbance of the solution: A = 0.995. The molar extinction coefficient of the product, ε, is 99.5 x 10^3 L mol^-1 cm^-1 and the path length of the cuvette is 1.0 cm. Calculate the concentration substrate after 100 s. Read more

Steps to Solve

1. Sketching [S] vs. time for zero order reaction

For a zero-order reaction, the rate of disappearance of the substrate is constant, meaning $[S]$ decreases linearly with time. Since $[P]$ increases linearly, $[S]$ must decrease linearly. If we assume an initial concentration of $[S]_0$ at $t=0$, $[S]$ will decrease according to $[S] = [S]_0 - kt$, where $k$ is the rate constant. Thus, plot a straight line with a negative slope such that $[S]$ decreases as $[P]$ increases.

2. Sketching [S] vs. time for first order reaction

For a first-order reaction, the rate of disappearance of the substrate is proportional to the substrate concentration. This means $[S]$ decreases exponentially with time. The equation is $[S] = [S]_0 e^{-kt}$. In this case, the rate of production of product P decreases over time, and thus the rate of S consumption also decreases over time. The plot of [S] concentration should represent an exponential decay.

3. Calculating the initial concentration of substrate [S]

First, calculate the initial concentration of the substrate after it has been added to the reaction mixture. We can use the dilution formula:

$C_1V_1 = C_2V_2$

where $C_1$ is the initial concentration of the substrate, $V_1$ is the volume of the substrate added, $C_2$ is the final concentration of the substrate in the reaction mixture, and $V_2$ is the total volume of the reaction mixture. Solving for $C_2$:

$C_2 = \frac{C_1V_1}{V_2} = \frac{(0.55 \text{ mM})(100 \text{ µL})}{2.3 \text{ mL}}$

4. Convert units

We convert the volumes to the same units. Since 1 mL = 1000 µL:

$C_2 = \frac{(0.55 \text{ mM})(100 \text{ µL})}{2300 \text{ µL}} = 0.0239 \text{ mM}$

5. Calculating the concentration of product [P] after 100 seconds using Beer-Lambert Law

We use the Beer-Lambert Law to find the concentration of the product after 100 seconds:

$A = \epsilon \cdot l \cdot [P]$ where $A$ is the absorbance, $\epsilon$ is the molar extinction coefficient, $l$ is the path length, and $[P]$ is the concentration of the product. Solving for $[P]$:

$[P] = \frac{A}{\epsilon \cdot l} = \frac{0.995}{(99.5 \times 10^3 \text{ L mol}^{-1} \text{cm}^{-1})(1.0 \text{ cm})}$

$[P] = 1.0 \times 10^{-5} \text{ M} = 0.01 \text{ mM}$

6. Calculating the concentration of substrate [S] after 100 seconds

Since the reaction is $S \rightarrow P$, the amount of substrate that has reacted is equal to the amount of product formed. Therefore, the concentration of substrate remaining after 100 seconds is:

$[S]_{100} = [S]0 - [P]{100} $

$[S]_{100} = 0.0239 \text{ mM} - 0.01 \text{ mM} = 0.0139 \text{ mM}$