Rate Constant: MDCAT Chemistry notes
Rate Constant MDCAT notes: specific rate constant k, what it depends on, units for each order, the Arrhenius equation and its graph, and half-life sums.
What the rate constant is
In the rate law rate $= k[\mathrm{A}]^x[\mathrm{B}]^y$, $k$ is the rate constant. When every reactant concentration is $1\ \mathrm{mol\,dm^{-3}}$, rate $= k$. For this reason $k$ is called the specific rate constant: it is the rate per unit concentration of reactants.
What k depends on
- Temperature: $k$ increases with temperature.
- Catalyst: a catalyst increases $k$ by lowering $E_a$.
- Nature of the reaction (its activation energy).
- $k$ is independent of the concentration of reactants. Changing concentration changes the rate, not $k$.
Units of k
General unit: $(\mathrm{mol\,dm^{-3}})^{1-n}\,\mathrm{s^{-1}}$ for order $n$.
| Order | Unit of $k$ |
|---|---|
| Zero | $\mathrm{mol\,dm^{-3}\,s^{-1}}$ (same as rate) |
| First | $\mathrm{s^{-1}}$ |
| Second | $\mathrm{dm^3\,mol^{-1}\,s^{-1}}$ |
| Third | $\mathrm{dm^6\,mol^{-2}\,s^{-1}}$ |
Only for a zero-order reaction does $k$ have the same unit as the rate, because rate $= k$.
Arrhenius equation
The effect of temperature on the rate constant is given by the Arrhenius equation:
$$k = Ae^{-E_a/RT}$$
- $A$ = Arrhenius (frequency) factor, $E_a$ = activation energy, $R$ = gas constant ($8.314\ \mathrm{J\,K^{-1}\,mol^{-1}}$), $T$ = absolute temperature.
- Higher $T$ or lower $E_a$ gives a larger $k$.
- Because $T$ appears in an exponent, even a small rise in temperature increases $k$ sharply; the effect is larger for reactions with a high $E_a$.
Taking logarithms gives the equation of a straight line:
$$\ln k = \ln A - \frac{E_a}{R}\cdot\frac{1}{T}\qquad \text{or}\qquad \log k = \log A - \frac{E_a}{2.303R}\cdot\frac{1}{T}$$
- Plot $\ln k$ against $1/T$: slope $= -E_a/R$, intercept $= \ln A$.
- Unit of slope: $\dfrac{\mathrm{J\,mol^{-1}}}{\mathrm{J\,K^{-1}\,mol^{-1}}} = \mathrm{K}$.
- From the slope, $E_a = -\text{slope}\times R$ (or $-\text{slope}\times 2.303R$ for the log plot).
Worked examples
Rate from k
A first-order reaction has $k = 0.020\ \mathrm{s^{-1}}$ and $[\mathrm{A}] = 0.50\ \mathrm{mol\,dm^{-3}}$. Rate $= k[\mathrm{A}] = 0.020\times0.50 = 0.010\ \mathrm{mol\,dm^{-3}\,s^{-1}}$.
Half-life from k
For a first-order reaction with $k = 0.0693\ \mathrm{min^{-1}}$: $t_{1/2} = \dfrac{0.693}{k} = \dfrac{0.693}{0.0693} = 10\ \mathrm{min}$.
Finding the unit
Rate $= k[\mathrm{A}]^2$. Then $k = \dfrac{\mathrm{mol\,dm^{-3}\,s^{-1}}}{(\mathrm{mol\,dm^{-3}})^2} = \mathrm{dm^3\,mol^{-1}\,s^{-1}}$.
Key formulas
- Rate $= k[\mathrm{A}]^x[\mathrm{B}]^y$
- $k = Ae^{-E_a/RT}$; $\ln k = \ln A - E_a/RT$
- First order: $t_{1/2} = 0.693/k$
Common MDCAT traps
- $k$ does not depend on concentration; it depends on temperature and catalyst.
- The exponent is $-E_a/RT$, with $R$ the gas constant.
- The Arrhenius plot ($\ln k$ vs $1/T$) is a straight line, not a curve.
- The unit of the slope is kelvin.
- First-order $k$ has unit $\mathrm{s^{-1}}$; zero-order $k$ has the same unit as rate.
Quick revision
- $k$ = rate at unit concentrations.
- Arrhenius equation relates $k$ to temperature.
- Slope of $\ln k$ vs $1/T$ is $-E_a/R$.
- Second-order $k$: $\mathrm{dm^3\,mol^{-1}\,s^{-1}}$.
- If $k = 0.693\ \mathrm{min^{-1}}$, first-order $t_{1/2}$ is 1 min.