Define instantaneous rate of reaction. Write down the rate expressions of the following reaction. 2N2O5(g) 4NO2(g) + O2(g) If 2.24 litres of O2 at NTP is produced in 30 minutes, what is the rate of…

Instantaneous Rate of Reaction

The instantaneous rate of reaction is the rate of change in concentration of a reactant or product at a specific instant of time during the reaction. It is determined by taking the derivative of concentration with respect to time at that particular moment. Mathematically, for a reactant $A$ or product $B$, it is expressed as:

$$\text{Rate}=-\frac{d[A]}{dt}\text{(for reactant)}\text{or}\text{Rate}=\frac{d[B]}{dt}\text{(for product)}$$

The negative sign for reactants indicates that their concentration decreases over time.

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Rate Expression for the Given Reaction

The reaction is: $$2\mathrm{N}{}_2\mathrm{O}{}_5(\mathrm{g})\to 4\mathrm{NO}{}_2(\mathrm{g})+\mathrm{O}{}_2(\mathrm{g})$$

The rate of reaction can be expressed in terms of the disappearance of $\mathrm{N}{}_2\mathrm{O}{}_5$ or the appearance of $\mathrm{NO}{}_2$ or $\mathrm{O}{}_2$. According to stoichiometry:

$$\text{Rate}=-\frac{1}{2}\frac{d[\mathrm{N}{}_2\mathrm{O}{}_5]}{dt}=\frac{1}{4}\frac{d[\mathrm{NO}{}_2]}{dt}=\frac{d[\mathrm{O}{}_2]}{dt}$$

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Calculation of the Rate of Disappearance of $\mathrm{N}{}_2\mathrm{O}{}_5$

Given:

• Volume of $\mathrm{O}{}_2$ produced = 2.24 L at NTP
• Time taken = 30 minutes = 0.5 hours

Step 1: Convert Volume to Moles

At NTP (Normal Temperature and Pressure), 1 mole of any gas occupies 22.4 L. Thus, moles of $\mathrm{O}{}_2$ produced:

$$n(\mathrm{O}{}_2)=\frac{2.24\text{ L}}{22.4{\text{ L mol}}^{-1}}=0.1\text{ mol}$$

Step 2: Determine Rate of Formation of $\mathrm{O}{}_2$

The rate of formation of $\mathrm{O}{}_2$ is:

$$\text{Rate of formation of }\mathrm{O}{}_2=\frac{\text{Moles of }\mathrm{O}{}_2}{\text{Time}}=\frac{0.1\text{ mol}}{0.5\text{ h}}=0.2{\text{ mol h}}^{-1}$$

Since the rate of reaction is equal to the rate of formation of $\mathrm{O}{}_2$ (from the rate expression above):

$$\text{Rate}=\frac{d[\mathrm{O}{}_2]}{dt}=0.2{\text{ mol h}}^{-1}$$

Step 3: Relate to the Rate of Disappearance of $\mathrm{N}{}_2\mathrm{O}{}_5$

From the rate expression:

$$\text{Rate}=-\frac{1}{2}\frac{d[\mathrm{N}{}_2\mathrm{O}{}_5]}{dt}=0.2{\text{ mol h}}^{-1}$$

Solving for the rate of disappearance of $\mathrm{N}{}_2\mathrm{O}{}_5$:

$$-\frac{d[\mathrm{N}{}_2\mathrm{O}{}_5]}{dt}=2\times 0.2=0.4{\text{ mol h}}^{-1}$$

Final Answer:

The rate of disappearance of $\mathrm{N}{}_2\mathrm{O}{}_5$ is $\boxed{0.4{\text{ mol h}}^{-1}}$.