Rate Constant (k)
Why k matters
The rate constant \(k\) is the key kinetic parameter. It defines how rapidly concentration decays for a selected reaction order.
\[ -\frac{dc}{dt}=k\,c^n \]
Larger \(k\) means faster conversion. The physical unit of \(k\) depends on reaction order \(n\).
- 0th order: concentration/time
- 1st order: 1/time
- 2nd order: 1/(concentration·time)
Formulas (MathJax)
\[ k_{(0)}=\frac{c_0-c(t)}{t},\quad k_{(1)}=\frac{\ln(c_0/c(t))}{t},\quad k_{(2)}=\frac{\frac{1}{c(t)}-\frac{1}{c_0}}{t} \]
\[ c(t)_{(0)}=c_0-k t,\quad c(t)_{(1)}=c_0\,e^{-kt},\quad c(t)_{(2)}=\frac{c_0}{1+k c_0 t} \]
\[ k_{(1)}=\frac{\ln 2}{t_{1/2}},\quad k_{(0)}=\frac{c_0}{2t_{1/2}},\quad k_{(2)}=\frac{1}{c_0 t_{1/2}} \]
Symbol legend
- \(k\): rate constant
- \(n\): reaction order
- \(c_0\): initial concentration
- \(c(t)\): concentration at time \(t\)
- \(t\): reaction time
- \(t_{1/2}\): half-life
- \(\ln\): natural logarithm
Detailed examples
Example 1 (1st order):
\(c_0=1.0\), \(c(t)=0.5\), \(t=4.62\)
\(k=\ln(1/0.5)/4.62\approx0.150\)
\(c_0=1.0\), \(c(t)=0.5\), \(t=4.62\)
\(k=\ln(1/0.5)/4.62\approx0.150\)
Example 2 (0th order):
\(c_0=1.2\), \(c(t)=0.6\), \(t=6\)
\(k=(1.2-0.6)/6=0.1\)
\(c_0=1.2\), \(c(t)=0.6\), \(t=6\)
\(k=(1.2-0.6)/6=0.1\)
Example 3 (2nd order):
\(c_0=1.0\), \(c(t)=0.4\), \(t=3\)
\(k=(1/0.4-1/1.0)/3=0.5\)
\(c_0=1.0\), \(c(t)=0.4\), \(t=3\)
\(k=(1/0.4-1/1.0)/3=0.5\)
Practical note:
Use the correct order model. A wrong \(n\) can still produce a numeric \(k\), but lead to poor prediction quality.
Use the correct order model. A wrong \(n\) can still produce a numeric \(k\), but lead to poor prediction quality.
Practical depth
Unit consistency
For 1st-order kinetics, \(k\) often uses \(\mathrm{s^{-1}}\) or \(\mathrm{min^{-1}}\). For 0th and 2nd order, units depend on the concentration unit used in the experiment.
Laboratory workflows
Estimate \(k\) from multiple data points and compare linearized representations. This reduces random measurement noise and improves kinetic confidence.
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