Question

Express the rate of the reaction $$ 2 \mathrm{~N}_{2} \mathrm{H}_{4}(l)+\mathrm{N}_{2} \mathrm{O}_{4}(l) \longrightarrow 3 \mathrm{~N}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g) $$ in terms of (a) \(\Delta\left[\mathrm{N}_{2} \mathrm{O}_{4}\right]\) (b) \(\Delta\left[\mathrm{N}_{2}\right]\)

Short answer

Question: Express the rate of the given reaction in terms of the change in concentrations of N2O4 and N2. Answer: (a) $$\frac{\Delta\left[\mathrm{N}_{2} \mathrm{O}_{4}\right]}{\Delta t} = -\frac{1}{2} \cdot \frac{\Delta\left[\mathrm{N}_{2}\mathrm{H}_{4}\right]}{\Delta t}$$, (b) $$\frac{\Delta\left[\mathrm{N}_{2}\right]}{\Delta t} = \frac{3}{1} \cdot \frac{\Delta\left[\mathrm{N}_{2}\mathrm{O}_{4}\right]}{\Delta t}$$.

Step-by-step solution

A. Rate law expression for N2O4

To express the rate of reaction in terms of $$\Delta\left[\mathrm{N}_{2} \mathrm{O}_{4}\right]$$, we will first write the general rate law for this reaction: Rate = k[2N2H4][N2O4] Now let's focus on the N2O4 species. Since it is a reactant, the rate expression for it will have a negative sign (because its concentration decreases during the reaction). Also, we need to consider the stoichiometry of the reaction, which states that one mole of N2O4 reacts with two moles of N2H4. So, for every mole of N2O4 that reacts, two moles of N2H4 are consumed. Therefore, the rate expression for N2O4 is: $$\frac{\Delta\left[\mathrm{N}_{2} \mathrm{O}_{4}\right]}{\Delta t} = -\frac{1}{2} \cdot \frac{\Delta\left[\mathrm{N}_{2}\mathrm{H}_{4}\right]}{\Delta t}$$.

B. Rate law expression for N2

Now let's express the rate of reaction in terms of $$\Delta\left[\mathrm{N}_{2}\right]$$. Since N2 is a product, the rate expression for it will have a positive sign (because its concentration increases during the reaction). Again, we need to consider the stoichiometry of the reaction, which states that three moles of N2 are produced for each mole of N2O4 reacting. So: $$\frac{\Delta\left[\mathrm{N}_{2}\right]}{\Delta t} = \frac{3}{1} \cdot \frac{\Delta\left[\mathrm{N}_{2}\mathrm{O}_{4}\right]}{\Delta t}$$. Now we have the rate expressions connected to the change in concentrations of both, N2O4 and N2, as asked in the exercise: (a) $$\frac{\Delta\left[\mathrm{N}_{2} \mathrm{O}_{4}\right]}{\Delta t} = -\frac{1}{2} \cdot \frac{\Delta\left[\mathrm{N}_{2}\mathrm{H}_{4}\right]}{\Delta t}$$ (b) $$\frac{\Delta\left[\mathrm{N}_{2}\right]}{\Delta t} = \frac{3}{1} \cdot \frac{\Delta\left[\mathrm{N}_{2}\mathrm{O}_{4}\right]}{\Delta t}$$.