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Question: The decomposition of N$_2$O$_5$ in CCl$_4$ at 318K has been studied by monitoring the concentration ...

The decomposition of N2_2O5_5 in CCl4_4 at 318K has been studied by monitoring the concentration of N2_2O5_5 in the solution. Initially the concentration of N2_2O5_5 is 2.33 mol L−1^{-1} and after 184 minutes, it is reduced to 2.08 mol L−1^{-1}. The reaction takes place by first order kinetics according to the equation:

2N2_2O5_5 (g) →\rightarrow 4NO2_2(g) + O2_2(g)

then select correct statement/s from the following - [consider log 1.12 = 0.05]

A

Average rate of reaction is 6.79×10−4\times10^{-4} mol L−1^{-1}/min

B

Average rate of formation of NO2(g)_{2(g)} is 2.72×10−3\times10^{-3} mol L−1^{-1} min−1^{-1}

C

Average rate of formation of O2(g)_{2(g)} is 1.36×10−3\times10^{-3} mol L−1^{-1} min−1^{-1}

D

Molecularity of the reaction is 2

Answer

Average rate of reaction is 6.79×10−4\times10^{-4} mol L−1^{-1}/min, Average rate of formation of NO2(g)_{2(g)} is 2.72×10−3\times10^{-3} mol L−1^{-1} min−1^{-1}

Explanation

Solution

The reaction is given by: 2N2_2O5_5 (g) →\rightarrow 4NO2_2(g) + O2_2(g)

Initial concentration of N2_2O5_5, [N2_2O5_5]initial_{initial} = 2.33 mol L−1^{-1} Final concentration of N2_2O5_5, [N2_2O5_5]final_{final} = 2.08 mol L−1^{-1} Time interval, Δt\Delta t = 184 minutes

The change in concentration of N2_2O5_5 is Δ[N2O5]=[N2O5]final−[N2O5]initial=2.08−2.33=−0.25\Delta[\text{N}_2\text{O}_5] = [\text{N}_2\text{O}_5]_{final} - [\text{N}_2\text{O}_5]_{initial} = 2.08 - 2.33 = -0.25 mol L−1^{-1}.

The average rate of disappearance of N2_2O5_5 is −Δ[N2O5]Δt=−−0.25184=0.25184-\frac{\Delta[\text{N}_2\text{O}_5]}{\Delta t} = -\frac{-0.25}{184} = \frac{0.25}{184} mol L−1^{-1} min−1^{-1}.

The average rate of reaction is defined as: Rate =−12Δ[N2O5]Δt=14Δ[NO2]Δt=11Δ[O2]Δt= -\frac{1}{2} \frac{\Delta[\text{N}_2\text{O}_5]}{\Delta t} = \frac{1}{4} \frac{\Delta[\text{NO}_2]}{\Delta t} = \frac{1}{1} \frac{\Delta[\text{O}_2]}{\Delta t}

Let's evaluate each statement:

Statement 1: Average rate of reaction is 6.79×10−4\times10^{-4} mol L−1^{-1}/min Average rate of reaction =12×(−Δ[N2O5]Δt)=12×0.25184=0.125184= \frac{1}{2} \times \left( -\frac{\Delta[\text{N}_2\text{O}_5]}{\Delta t} \right) = \frac{1}{2} \times \frac{0.25}{184} = \frac{0.125}{184} mol L−1^{-1} min−1^{-1}. Calculating the value: 0.125184≈0.0006793478\frac{0.125}{184} \approx 0.0006793478 mol L−1^{-1} min−1^{-1}. This is approximately 6.79×10−46.79 \times 10^{-4} mol L−1^{-1} min−1^{-1}. Statement 1 is correct.

Statement 2: Average rate of formation of NO2(g)_{2(g)} is 2.72×10−3\times10^{-3} mol L−1^{-1} min−1^{-1} The average rate of formation of NO2_2 is Δ[NO2]Δt\frac{\Delta[\text{NO}_2]}{\Delta t}. From the stoichiometry, 14Δ[NO2]Δt=−12Δ[N2O5]Δt\frac{1}{4} \frac{\Delta[\text{NO}_2]}{\Delta t} = -\frac{1}{2} \frac{\Delta[\text{N}_2\text{O}_5]}{\Delta t}. So, Δ[NO2]Δt=−4×12Δ[N2O5]Δt=−2Δ[N2O5]Δt=2×(−Δ[N2O5]Δt)\frac{\Delta[\text{NO}_2]}{\Delta t} = -4 \times \frac{1}{2} \frac{\Delta[\text{N}_2\text{O}_5]}{\Delta t} = -2 \frac{\Delta[\text{N}_2\text{O}_5]}{\Delta t} = 2 \times \left( -\frac{\Delta[\text{N}_2\text{O}_5]}{\Delta t} \right). Δ[NO2]Δt=2×0.25184=0.50184\frac{\Delta[\text{NO}_2]}{\Delta t} = 2 \times \frac{0.25}{184} = \frac{0.50}{184} mol L−1^{-1} min−1^{-1}. Calculating the value: 0.50184≈0.00271739\frac{0.50}{184} \approx 0.00271739 mol L−1^{-1} min−1^{-1}. This is approximately 2.72×10−32.72 \times 10^{-3} mol L−1^{-1} min−1^{-1}. Statement 2 is correct.

Statement 3: Average rate of formation of O2(g)_{2(g)} is 1.36×10−3\times10^{-3} mol L−1^{-1} min−1^{-1} The average rate of formation of O2_2 is Δ[O2]Δt\frac{\Delta[\text{O}_2]}{\Delta t}. From the stoichiometry, 11Δ[O2]Δt=−12Δ[N2O5]Δt\frac{1}{1} \frac{\Delta[\text{O}_2]}{\Delta t} = -\frac{1}{2} \frac{\Delta[\text{N}_2\text{O}_5]}{\Delta t}. So, Δ[O2]Δt=12×(−Δ[N2O5]Δt)\frac{\Delta[\text{O}_2]}{\Delta t} = \frac{1}{2} \times \left( -\frac{\Delta[\text{N}_2\text{O}_5]}{\Delta t} \right). Δ[O2]Δt=12×0.25184=0.125184\frac{\Delta[\text{O}_2]}{\Delta t} = \frac{1}{2} \times \frac{0.25}{184} = \frac{0.125}{184} mol L−1^{-1} min−1^{-1}. Calculating the value: 0.125184≈0.0006793478\frac{0.125}{184} \approx 0.0006793478 mol L−1^{-1} min−1^{-1}. This is approximately 6.79×10−46.79 \times 10^{-4} mol L−1^{-1} min−1^{-1}. The statement claims the rate is 1.36×10−31.36 \times 10^{-3} mol L−1^{-1} min−1^{-1}, which is approximately twice our calculated value. Statement 3 is incorrect.

Statement 4: Molecularity of the reaction is 2 Molecularity is the number of reactant species (atoms, molecules, or ions) that collide simultaneously in an elementary reaction step. It is a theoretical concept applicable only to elementary reactions. The given equation 2N2_2O5_5 (g) →\rightarrow 4NO2_2(g) + O2_2(g) represents the overall stoichiometry of the reaction, not necessarily an elementary step. The problem states that the reaction follows first-order kinetics. This implies that the rate law is Rate = k[N2O5]k[\text{N}_2\text{O}_5]. A reaction with molecularity 2 (bimolecular) would typically have a rate law that depends on the concentration of two species (or the square of the concentration of one species). Since the reaction is first order, it is a complex reaction consisting of multiple elementary steps, and the rate-determining step likely involves one molecule or leads to an overall first-order dependence on N2_2O5_5. Molecularity is not defined for a complex reaction as a whole. Therefore, stating the molecularity of this overall reaction is 2 is incorrect. Statement 4 is incorrect.

Based on the calculations, statements 1 and 2 are correct.