Problem 60 - General Chemistry: Principles and Modern Applications

Question

The following substrate concentration [S] versus time data were obtained during an enzyme-catalyzed reaction: \(t=0 \min ,[\mathrm{S}]=1.00 \mathrm{M} ; 20 \mathrm{min}, 0.90 \mathrm{M}; 60 \min , 0.70 \mathrm{M} ; 100 \mathrm{min}, 0.50 \mathrm{M} ; 160 \mathrm{min}, 0.20 \mathrm{M}.\) What is the order of this reaction with respect to \(\mathrm{S}\) in the concentration range studied?

Step-by-Step Solution

Answer

The reaction is first order with respect to S.

Step 1: Identify the type of reaction

The types of reactions typically studied in kinetics are zero order, first order, and second order. We know a reaction is zero order if the rate is independent of the concentration of the reactant. The reaction is first order if the rate is directly proportional to the concentration of the reactant. It's a second order reaction if the rate is proportional to the square of the concentration of the reactant. Based on these principles, we will check for each type of reaction.

Step 2: Checking if it's a zero order reaction

In a zero order reaction, concentration [S] will decrease linearly with time. From the provided set of data, we see that the concentration is not decreasing linearly over time, hence it is not a zero order reaction.

Step 3: Checking if it's a first order reaction

In a first order reaction, assuming \([S]\) at \(t=0\) is \(S_0\), and at time \(t\) is \([S]\), we would use the equation, \(\ln([S]/S_0) = -kt\). Choose two data points, for example, \((t=0 min, [S]= 1.00M)\) and \((t=20min, [S]=0.90M)\) , if we substitute these in our equation, \(\ln{(0.90/1)} = -k * 20\) and solve for \(k\), repeat this for multiple pairs and if \(k\) value is nearly constant we can say it is a first order reaction.

Step 4: Checking if it's a second order reaction

The approach is similar to the first order reaction but with a different equation: \(1/[S] - 1/S_0 = kt\). If after substituting pairs of \([S]\) and \(t\) in above equation, we get nearly constant \(k\), it's a second order reaction. However, given previous analysis we wouldn't need to do this calculation.

Key Concepts

Reaction Order

The concept of reaction order is foundational in understanding how the rate of a chemical reaction depends on the concentration of the reactants. Essentially, the order tells us how the speed of the reaction changes as the concentration of a reactant changes.
The order of reaction can be determined experimentally and usually takes values like zero, one, or two, corresponding to zero-order, first-order, and second-order reactions. Each of these has distinct characteristics and mathematical representations.

• Zero-order reactions have a constant rate, independent of the concentration of the reactant.
• First-order reactions have a rate directly proportional to reactant concentration.
• Second-order reactions have a rate proportional to the square of the concentration of the reactant.

Knowing the reaction order helps predict how changes in concentration will affect the reaction rate, which is crucial in fields like pharmacology, environmental science, and many industrial processes.

Zero Order Reaction

In a zero-order reaction, the rate of reaction is constant. This means the reaction proceeds at a steady rate, regardless of how much reactant is present. Because the rate is independent of the concentration of the reactant, the reaction can only proceed until the reactant is depleted.
A characteristic equation of a zero-order reaction is:\[ [S] = [S_0] - kt \]where

• \([S]\) is the concentration of the substrate at time \(t\),
• \([S_0]\) is the initial concentration, and
• \(k\) is the zero-order rate constant.

The graph of concentration vs. time for a zero-order reaction is a straight line, decreasing linearly. In practice, zero-order reactions are rare and usually occur in specific conditions, such as when an enzyme is saturated with substrate.

First Order Reaction

A first-order reaction is one where the rate depends linearly on the concentration of one reactant. As the concentration decreases, the rate decreases proportionately. This is common in many decay processes, such as radioactive decay or enzyme catalysis when the enzyme is not saturated.
The mathematical equation for a first-order reaction is:\[ \ln([S]/S_0) = -kt\]Here:

• \([S]\) represents the concentration at time \(t\),
• \([S_0]\) is the initial concentration, and
• \(k\) is the first-order rate constant.

When plotted, the natural logarithm of concentration versus time will give a straight line, allowing for easy determination of the rate constant, \(k\). Understanding this can help in interpreting various natural and industrial processes where reactant concentration directly affects reaction rate.

Second Order Reaction

Second-order reactions can occur in two scenarios: one where the reaction depends on the concentration of two different reactants, or one where it depends on the square of the concentration of a single reactant. Such reactions typically show a rapid initial change in reactant concentration, which gradually slows down.
The formula for a second-order reaction is:\[ 1/[S] - 1/S_0 = kt \]where:

• \([S]\) is the concentration at time \(t\),
• \([S_0]\) is initial concentration, and
• \(k\) is the second-order rate constant.

When this data is plotted as the inverse of concentration vs. time, it should yield a straight line. These reactions are common in situations where molecules must collide for a reaction to occur, such as in bimolecular reactions.