NO2 required for a reaction is produced by the decomposition of N2O5 in CCl4 as per the equation, 2N2O5(g) → 4NO2(g) + O2(g). The initial concentration of N2O5 is 3.00 mol L–1 and it is 2.75 mol L–1 after 30 minutes. The rate of formation of NO2 is :

Engineering

Chemistry

Rate of Chemical Reaction

Integrated Rate Law

Question

NO₂ required for a reaction is produced by the decomposition of N₂O₅ in CCl₄ as per the equation,

2N₂O₅(g) → 4NO₂(g) + O₂(g).

The initial concentration of N₂O₅ is 3.00 mol L^–1 and it is 2.75 mol L^–1 after 30 minutes. The rate of formation of NO₂ is :

2.083 × 10^–3 mol L^–1 min^–1

8.333 × 10^–3 mol L^–1 min^–1

1.667 × 10^–2 mol L^–1 min^–1

4.167 × 10^–3 mol L^–1 min^–1

$\frac{1}{2} \frac{d [N_{2} O_{5}]}{dt} = - \frac{1}{4} \frac{d [{NO}_{2}]}{dt}$

$\Rightarrow \frac{d [{NO}_{2}]}{dt} = - 2 \frac{d [N_{2} O_{5}]}{dt}$

$\Rightarrow \frac{d [{NO}_{2}]}{dt} = - 2 \frac{(3 - 2 .75)}{30}$

$\frac{d [{NO}_{2}]}{dt} = \frac{2 \times 0 .25}{30} = 1 .667 \times 10^{2} mol L^{- 1} \min^{- 1}$

Understanding the Problem

We are given the reaction: $2 N_{2} O_{5} (g) \to 4 N O_{2} (g) + O_{2} (g)$ .

Initial concentration of N₂O₅, [N₂O₅]₀ = 3.00 mol L^–1.

Concentration after 30 minutes, [N₂O₅]_t = 2.75 mol L^–1.

We need to find the rate of formation of NO₂.

Step-by-Step Solution

Step 1: Find the Rate of Disappearance of N₂O₅

The average rate of disappearance of a reactant is given by:

$Rate = - \frac{1}{2} \frac{Δ [N_{2} O_{5}]}{Δ t}$

First, calculate the change in concentration, $Δ [N_{2} O_{5}] = [N_{2} O_{5}] \to_{t} - [N_{2} O_{5}] \to_{0} = 2.75 mol/L - 3.00 mol/L = - 0.25 mol/L$

Change in time, $Δ t = 30 min$

Now, calculate the average rate of disappearance of N₂O₅:

$Rate = - \frac{1}{2} \frac{(- 0.25 mol/L)}{30 min} = - \frac{1}{2} \times (- 0.008333... mol L^{- 1} {min}^{- 1}) = 0.0041667 mol L^{- 1} {min}^{- 1}$

So, the average rate of the reaction is approximately $4.167 \times 10^{- 3} mol L^{- 1} {min}^{- 1}$ .

Step 2: Relate the Reaction Rate to the Rate of Formation of NO₂

The rate of a reaction can be expressed in terms of any reactant or product using their stoichiometric coefficients.

The general relationship is: $Rate = - \frac{1}{2} \frac{d [N_{2} O_{5}]}{d t} = + \frac{1}{4} \frac{d [N O_{2}]}{d t}$

Therefore, the rate of formation of NO₂ is:

$\frac{d [N O_{2}]}{d t} = 4 \times Rate$

We substitute the value of the reaction rate we calculated:

$\frac{d [N O_{2}]}{d t} = 4 \times (4.167 \times 10^{- 3} mol L^{- 1} {min}^{- 1})$

$\frac{d [N O_{2}]}{d t} = 0.016668 mol L^{- 1} {min}^{- 1} = 1.667 \times 10^{- 2} mol L^{- 1} {min}^{- 1}$

Final Answer

The rate of formation of NO₂ is $1.667 \times 10^{- 2} mol L^{- 1} {min}^{- 1}$ .

Experiment	[A] (inmolL^-1)	[B] (inmolL^-1)	Initilalrate of reaction (in mol L^–1min^–1)
I	0.10	0.20	6.93 × 10^-3
II	0.10	0.25	6.93 × 10^-3
III	0.20	0.30	1.386 × 10^-2

Detailed Solution

Understanding the Problem

Step-by-Step Solution

Step 1: Find the Rate of Disappearance of N₂O₅

Step 2: Relate the Reaction Rate to the Rate of Formation of NO₂

Final Answer

Related Topics & Formulae

Key Concepts

Important Formulae

Detailed Solution

Understanding the Problem

Step-by-Step Solution

Step 1: Find the Rate of Disappearance of N2O5

Step 2: Relate the Reaction Rate to the Rate of Formation of NO2

Final Answer

Related Topics & Formulae

Key Concepts

Important Formulae

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Step 1: Find the Rate of Disappearance of N₂O₅

Step 2: Relate the Reaction Rate to the Rate of Formation of NO₂