Chapter 14: Problem 60
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?
Short Answer
Step by step solution
Identify the type of reaction
Checking if it's a zero order reaction
Checking if it's a first order reaction
Checking if it's a second order reaction
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Reaction Order
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.
Zero Order Reaction
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.
First Order Reaction
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.
Second Order Reaction
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.