What is the equation for a first rate reaction?
2 Answers
The equation for a first-order reaction can be expressed as:
\[
\text{Rate} = k[A]
\]
where:
- \(\text{Rate}\) is the rate of the reaction,
- \(k\) is the rate constant,
- \([A]\) is the concentration of the reactant.
For first-order reactions, the integrated rate law can also be written as:
\[
\ln[A] = -kt + \ln[A_0]
\]
where \([A_0]\) is the initial concentration of the reactant. This equation shows how the concentration of the reactant decreases over time.
\[
\text{Rate} = k[A]
\]
where:
- \(\text{Rate}\) is the rate of the reaction,
- \(k\) is the rate constant,
- \([A]\) is the concentration of the reactant.
For first-order reactions, the integrated rate law can also be written as:
\[
\ln[A] = -kt + \ln[A_0]
\]
where \([A_0]\) is the initial concentration of the reactant. This equation shows how the concentration of the reactant decreases over time.
A first-order reaction is a type of chemical reaction where the rate of reaction is directly proportional to the concentration of one reactant. The general form of the rate law for a first-order reaction is:
\[ \text{Rate} = k [A] \]
where:
- \(\text{Rate}\) is the reaction rate.
- \(k\) is the rate constant.
- \([A]\) is the concentration of the reactant A.
For a first-order reaction, the rate constant \(k\) has units of \(\text{time}^{-1}\), such as \(\text{s}^{-1}\) in seconds.
### Integrated Rate Law
The integrated rate law for a first-order reaction, which relates the concentration of the reactant to time, is:
\[ \ln[A]_t = -kt + \ln[A]_0 \]
where:
- \([A]_t\) is the concentration of the reactant at time \(t\).
- \([A]_0\) is the initial concentration of the reactant.
- \(k\) is the rate constant.
- \(t\) is the time elapsed.
Rearranging this equation to solve for the concentration at any time \(t\):
\[ [A]_t = [A]_0 e^{-kt} \]
### Half-Life
The half-life (\(t_{1/2}\)) of a first-order reaction, which is the time required for the concentration of the reactant to decrease to half of its initial value, is given by:
\[ t_{1/2} = \frac{\ln(2)}{k} \]
This equation shows that the half-life of a first-order reaction is constant and independent of the initial concentration.
### Example
If a reaction follows the rate law \( \text{Rate} = k [A] \) with a rate constant \( k = 0.1 \text{s}^{-1} \), and the initial concentration \([A]_0\) is 0.5 M, the concentration after 10 seconds can be calculated as:
\[ [A]_{10} = 0.5 \, \text{M} \times e^{-0.1 \times 10} \approx 0.5 \, \text{M} \times e^{-1} \approx 0.5 \, \text{M} \times 0.3679 \approx 0.184 \, \text{M} \]
\[ \text{Rate} = k [A] \]
where:
- \(\text{Rate}\) is the reaction rate.
- \(k\) is the rate constant.
- \([A]\) is the concentration of the reactant A.
For a first-order reaction, the rate constant \(k\) has units of \(\text{time}^{-1}\), such as \(\text{s}^{-1}\) in seconds.
### Integrated Rate Law
The integrated rate law for a first-order reaction, which relates the concentration of the reactant to time, is:
\[ \ln[A]_t = -kt + \ln[A]_0 \]
where:
- \([A]_t\) is the concentration of the reactant at time \(t\).
- \([A]_0\) is the initial concentration of the reactant.
- \(k\) is the rate constant.
- \(t\) is the time elapsed.
Rearranging this equation to solve for the concentration at any time \(t\):
\[ [A]_t = [A]_0 e^{-kt} \]
### Half-Life
The half-life (\(t_{1/2}\)) of a first-order reaction, which is the time required for the concentration of the reactant to decrease to half of its initial value, is given by:
\[ t_{1/2} = \frac{\ln(2)}{k} \]
This equation shows that the half-life of a first-order reaction is constant and independent of the initial concentration.
### Example
If a reaction follows the rate law \( \text{Rate} = k [A] \) with a rate constant \( k = 0.1 \text{s}^{-1} \), and the initial concentration \([A]_0\) is 0.5 M, the concentration after 10 seconds can be calculated as:
\[ [A]_{10} = 0.5 \, \text{M} \times e^{-0.1 \times 10} \approx 0.5 \, \text{M} \times e^{-1} \approx 0.5 \, \text{M} \times 0.3679 \approx 0.184 \, \text{M} \]