Rate of reaction
Consider a general reaction: aA + bB → cC + dD
where a, b, c and d are the stoichiometric coefficients of reactants and products. The rate expression for this reaction is Rate ∝ [A]x[B]y
where exponents x and y may or may not be equal to the stoichiometric coefficients (a and b) of the reactants. Above equation can also be written as Rate = k[A]x[B]y
Sum of these exponents, i.e., x + y gives the overall order of a reaction whereas x and y represent the order with respect to the reactants A and B, respectively.
\( \displaystyle -\frac{d[\mathrm{R}]}{dt}=k[\mathrm{A}]^x[\mathrm{B}]^y \)
The above equation is known as differential rate equation, where k is a proportionality constant called rate constant. The equation, which relates the rate of a reaction to concentration of reactants is called rate law or rate expression.
In the rate law expression reaction rate is given in terms of molar concentration of reactants with each term raised to some power, which may or may not be same as the stoichiometric coefficient of the reacting species in a balanced chemical equation. For example:
2NO2(g) + O2(g) → 2NO3(g)
The rate of this reaction is measured as a function of initial concentrations either by keeping the concentration of one of the reactants constant and changing the concentration of the other reactant or by changing the concentration of both the reactants. The results obtained are tabulated in table below:
| Experiment | Initial [NO2]/mol L−1 | Initial [O2]/mol L−1 | Initial rate of formation of [NO3]/mol L−1s−1 |
| 1 | 0.25 | 0.25 | 0.090 |
| 2 | 0.50 | 0.25 | 0.360 |
| 3 | 0.25 | 0.50 | 0.180 |
| 4 | 0.50 | 0.50 | 0.720 |
What is the order of the reaction with respect to NO2?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → Two
The order with respect to a reactant is found by the initial rate method: compare two experiments in which only that reactant’s concentration changes and all others stay the same. If its concentration is multiplied by a factor f and the rate is multiplied by fx, the order with respect to it is x.
For Rate = k\( \ce{[NO2]} \)x\( \ce{[O2]} \)y, take Experiments 1 and 2, where \( \ce{[O2]} \) is fixed at 0.25 mol L−1 and \( \ce{[NO2]} \) doubles from 0.25 to 0.50 mol L−1:
\( \displaystyle \frac{\text{Rate}_2}{\text{Rate}_1}=\frac{k(0.50)^x(0.25)^y}{k(0.25)^x(0.25)^y} \)
\( \displaystyle \frac{0.360}{0.090}=\left(\frac{0.50}{0.25}\right)^x \)
\( \displaystyle 4=2^x \quad\Rightarrow\quad x=2 \)
Doubling \( \ce{[NO2]} \) makes the reaction four times faster, so the order with respect to \( \ce{NO2} \) is two.
Note: It seems that there is a mistake in the original NTA paper, which prints 2NO2(g) + O2(g) → 2NO3(g) when it should read 2NO(g) + O2(g) → 2NO2(g).