Rate equation
The rate equation (or rate law) is an equation used to calculate the speed of a chemical reaction (or rate of reaction). For a general reaction aA + bB → C, the rate equation is:
- [math]\displaystyle{ v\; =\; k[\mathrm{A}]^x[\mathrm{B}]^y }[/math]
Here, [A] and [B] are the concentrations of A and B. x and y depend on which step is rate-determining. If the reaction mechanism is a very simple one, where A and B hit each other and then go to products through one transition state, then x=a and y=b. k is the rate constant of the reaction. This changes with temperature, pressure and other conditions.[1][1]
The rate equation is a differential equation. If it is integrated, then an equation which tells how the concentration of reagents and products changes with time is found.
In special cases, it is very easy to solve the equation and find k. For example, in a first-order reaction the equation is:
- [math]\displaystyle{ v = -\frac{d[A]}{dt} = k[A] }[/math]
Integrating gives:
- [math]\displaystyle{ \ \ln{[A]} = -kt + \ln{[A]_0} }[/math]
So, a plot of [math]\displaystyle{ \ln{[A]} }[/math] against time t gives a straight line with a slope of [math]\displaystyle{ -k }[/math].
Sometimes, experiments can be made so that the reaction looks like a first-order one. If the concentration of one reagent is kept at the same high value then it can be thought of as a constant. The equation becomes [math]\displaystyle{ v=k[A][B]=k'[A] }[/math] where k' is the pseudo-first order rate constant. Then the method above can be used to calculate k'.
Rate Equation Media
Concentration of A (A0 = 0.25 mol/L) and B versus time reaching equilibrium k1 = 2 min−1 and k−1 = 1 min−1
Time course of two first order, competitive reactions with differing rate constants.
