$\text{NO}_2 \text{ required for a reaction is produced by decomposition of } \text{N}_2\text{O}_5 \text{ in } \text{CCl}_4 \text{ as by equation}$
\[ 2\text{N}_2\text{O}_{5(g)} \rightarrow 4\text{NO}_{2(g)} + \text{O}_{2(g)} \]
$\text{The initial concentration of } \text{N}_2\text{O}_5 \text{ is } 3 \text{ mol L}^{-1} \text{ and it is } 2.75 \text{ mol L}^{-1} \text{ after 30 minutes.}$
$\text{The rate of formation of } \text{NO}_2 \text{ is } x \times 10^{-3} \text{ mol L}^{-1} \text{ min}^{-1} , \text{ value of } x \text{ is \_\_\_\_\_\_.}$
\[ 2\text{N}_2\text{O}_{5(g)} \rightarrow 4\text{NO}_{2(g)} + \text{O}_{2(g)} \]
$\text{The initial concentration of } \text{N}_2\text{O}_5 \text{ is } 3 \text{ mol L}^{-1} \text{ and it is } 2.75 \text{ mol L}^{-1} \text{ after 30 minutes.}$
$\text{The rate of formation of } \text{NO}_2 \text{ is } x \times 10^{-3} \text{ mol L}^{-1} \text{ min}^{-1} , \text{ value of } x \text{ is \_\_\_\_\_\_.}$
Correct Answer: 17
Approach Solution - 1
Solution: To find the rate of formation of \( NO_2 \), we first need to determine the change in concentration of \( N_2O_4 \) over the given time period.
Initial and Final Concentration:
Initial concentration of \( N_2O_4 \): \([N_2O_4]_0 = 3 \, \text{mol L}^{-1}\)
Final concentration after 30 minutes: \([N_2O_4] = 2.75 \, \text{mol L}^{-1}\)
Change in Concentration:
The change in concentration of \( N_2O_4 \) over 30 minutes is:
\[ \Delta [N_2O_4] = [N_2O_4]_0 - [N_2O_4] = 3 - 2.75 = 0.25 \, \text{mol L}^{-1} \]
Stoichiometry of the Reaction:
According to the reaction:
\[ 2N_2O_4 \rightarrow 4NO_2 \]
For every 2 moles of \( N_2O_4 \) that decompose, 4 moles of \( NO_2 \) are formed, so the ratio is:
\[ \frac{4 \, \text{mol} \, NO_2}{2 \, \text{mol} \, N_2O_4} = 2 \]
Rate of Formation of \( NO_2 \):
The change in concentration of \( NO_2 \) formed is:
\[ \Delta [NO_2] = 2 \times \Delta [N_2O_4] = 2 \times 0.25 = 0.50 \, \text{mol L}^{-1} \]
The rate of formation of \( NO_2 \) over 30 minutes is:
\[ \text{Rate} = \frac{\Delta [NO_2]}{\Delta t} = \frac{0.50 \, \text{mol L}^{-1}}{30 \, \text{min}} = \frac{0.50}{30} \, \text{mol L}^{-1} \text{min}^{-1} = \frac{1}{60} \, \text{mol L}^{-1} \text{min}^{-1} \]
Convert to \( x \):
Given \( x \times 10^{-3} = \frac{1}{60} \), we can find \( x \):
\[ x = \frac{1}{60} \times 1000 = 16.67 \approx 17 \]
Thus, the value of \( x \) is: 17
Approach Solution -2
Given the reaction \( 2\text{N}_2\text{O}_{5(g)} \rightarrow 4\text{NO}_{2(g)} + \text{O}_{2(g)} \), the initial concentration of \( \text{N}_2\text{O}_5 \) is \( 3 \, \text{mol L}^{-1} \) and it becomes \( 2.75 \, \text{mol L}^{-1} \) after 30 minutes. We need to find the rate of formation of \( \text{NO}_2 \) in the form \( x \times 10^{-3} \, \text{mol L}^{-1} \text{ min}^{-1} \) and determine the value of \( x \).
Concept Used:
The rate of a reaction can be expressed in terms of the disappearance of a reactant or the appearance of a product. For a general reaction \( aA \rightarrow bB \), the relationship between the rates is given by: \[ -\frac{1}{a} \frac{d[A]}{dt} = +\frac{1}{b} \frac{d[B]}{dt} \] In this case, the stoichiometry of the reaction \( 2\text{N}_2\text{O}_5 \rightarrow 4\text{NO}_2 \) links the rate of decomposition of \( \text{N}_2\text{O}_5 \) to the rate of formation of \( \text{NO}_2 \). The average rate over a time interval is calculated using the change in concentration.
Step-by-Step Solution:
Step 1: Calculate the average rate of decomposition of \( \text{N}_2\text{O}_5 \).
\[ \text{Average rate of decomposition of } \text{N}_2\text{O}_5 = -\frac{\Delta [\text{N}_2\text{O}_5]}{\Delta t} = -\frac{(2.75 - 3.00) \, \text{mol L}^{-1}}{30 \, \text{min}} \] \[ = -\frac{(-0.25) \, \text{mol L}^{-1}}{30 \, \text{min}} = \frac{0.25}{30} \, \text{mol L}^{-1} \text{ min}^{-1} = 8.333 \times 10^{-3} \, \text{mol L}^{-1} \text{ min}^{-1} \]
Step 2: Relate the rate of decomposition of \( \text{N}_2\text{O}_5 \) to the rate of formation of \( \text{NO}_2 \) using the reaction stoichiometry.
From the balanced equation \( 2\text{N}_2\text{O}_5 \rightarrow 4\text{NO}_2 \), the stoichiometric ratio is:
\[ \frac{\text{Rate of formation of NO}_2}{\text{Rate of decomposition of N}_2\text{O}_5} = \frac{4}{2} = 2 \]
Therefore:
\[ \text{Rate of formation of NO}_2 = 2 \times (\text{Rate of decomposition of N}_2\text{O}_5) \]
Step 3: Calculate the rate of formation of \( \text{NO}_2 \).
\[ \text{Rate of formation of NO}_2 = 2 \times (8.333 \times 10^{-3} \, \text{mol L}^{-1} \text{ min}^{-1}) = 16.666 \times 10^{-3} \, \text{mol L}^{-1} \text{ min}^{-1} \]
Step 4: Express the answer in the required form \( x \times 10^{-3} \) and find \( x \).
\[ \text{Rate of formation of NO}_2 = 16.666 \times 10^{-3} \approx 16.7 \times 10^{-3} \, \text{mol L}^{-1} \text{ min}^{-1} \]
Thus, the value of \( x \) is 17 (nearest integer).
Top JEE Main Chemistry Questions
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- At equilibrium the concentrations of and in a sealed vessel at . What will be for the reaction
- Find out the major products from the following reactions
Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are
- The correct order of atomic radii is:
Top JEE Main Chemical Kinetics Questions
- The rate of a first order reaction is \( 0.04 \, \text{mol} \, \text{L}^{-1} \, \text{s}^{-1} \) at 10 minutes and \( 0.03 \, \text{mol} \, \text{L}^{-1} \, \text{s}^{-1} \) at 20 minutes after initiation. Half-life of the reaction is \( \_\_\_\_ \) minutes. (Given \( \log 2 = 0.3010, \, \log 3 = 0.4771 \))
- $\text{Time required for completion of 99.9\% of a first order reaction is \_\_\_\_\_\_ times of half life } (t_{1/2}) \text{ of the reaction.}$
Consider the following data for the given reaction
\(2\)\(\text{HI}_{(g)}\) \(\rightarrow\) \(\text{H}_2{(g)}\)$ + $\(\text{I}_2{(g)}\)
The order of the reaction is __________.- For a reaction taking place in three steps at the same temperature, the overall rate constant \( K = \frac{K_1 K_2}{K_3} \). If \( E_{a1} \), \( E_{a2} \), and \( E_{a3} \) are 40, 50, and 60 kJ/mol respectively, the overall \( E_a \) is ____ kJ/mol.
For a reaction taking place in three steps at the same temperature, the overall rate constant \( K = \frac{K_1 K_2}{K_3} \). If \( E_{a1} \), \( E_{a2} \), and \( E_{a3} \) are 40, 50, and 60 kJ/mol respectively, the overall \( E_a \) is ____ kJ/mol.
Top JEE Main Questions
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,