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Chapter 7: Problem 140

One mole of \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g})\) at \(300 \mathrm{~K}\) is kept in a closed container under one atmospheric pressure. It is heated to \(600 \mathrm{~K}\) when \(20 \%\) by mass of \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g})\) decomposes to \(\mathrm{NO}_{2}(\mathrm{~g})\). The resultant pressure is (a) \(1.0 \mathrm{~atm}\) (b) \(2.4 \mathrm{~atm}\) (c) \(2.0 \mathrm{~atm}\) (d) \(1.2 \mathrm{~atm}\)

Short Answer

Expert verified
The resultant pressure is 2.4 atm.

Step by step solution

01

Initial Conditions Analysis

Initially, we have 1 mole of \(\mathrm{N}_2\mathrm{O}_4\). Given that the temperature is \(300\,\mathrm{K}\), the pressure in the container is \(1.0\,\mathrm{atm}\). The reaction that occurs upon heating is \(\mathrm{N}_2\mathrm{O}_4 (g) \rightarrow 2\mathrm{NO}_2 (g)\).
02

Decomposition Calculation

At \(600\,\mathrm{K}\), \(20\%\) of \(\mathrm{N}_2\mathrm{O}_4\) has decomposed. Therefore, \(0.2\, \text{moles of} \; \mathrm{N}_2\mathrm{O}_4\) decompose, forming \(0.4\, \text{moles of} \; \mathrm{NO}_2\) (since 1 mole of \(\mathrm{N}_2\mathrm{O}_4\) gives 2 moles of \(\mathrm{NO}_2\)).
03

Calculation of Total Moles After Reaction

Following the decomposition, \(0.8\) moles of \(\mathrm{N}_2\mathrm{O}_4\) remain and \(0.4\) moles of \(\mathrm{NO}_2\) are formed, making the total moles in the container \(0.8 + 0.4 = 1.2\, \text{moles}.\)
04

Application of Ideal Gas Law at New Conditions

Using the ideal gas law under the new conditions at \(600\,\mathrm{K}\), \[PV = nRT \Rightarrow P = \frac{nRT}{V} \] where \(V\) is constant. Let original pressure \(P_1 = 1\,\mathrm{atm}, n_1 = 1\,\text{mole}, T_1 = 300\,\mathrm{K}\) and we want to find \(P_2\) when \(n_2 = 1.2\,\text{moles}, T_2 = 600\,\mathrm{K}\):\[\frac{P_1}{P_2} = \frac{n_1T_1}{n_2T_2} \Rightarrow P_2 = P_1 \times \frac{n_2T_2}{n_1T_1} = 1 \times \frac{1.2 \times 600}{1 \times 300} = 2.4\,\mathrm{atm}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Equilibrium
Chemical equilibrium is the final state in a chemical reaction where the rate of the forward reaction equals the rate of the reverse reaction. This means that the concentrations of all reactants and products remain constant over time. In the provided exercise,
  • We have an initial equilibrium established with only one compound, \( \mathrm{N}_2\mathrm{O}_4 (g) \), at 300 K and 1 atm.

  • Upon heating the system to 600 K, some of the \( \mathrm{N}_2\mathrm{O}_4 \) decomposes to form \( \mathrm{NO}_2 \) gas, a new state of equilibrium is reached as the system tries to adjust to this change.

The reaction \[\mathrm{N}_2\mathrm{O}_4 (g) \rightleftharpoons 2\mathrm{NO}_2 (g)\]illustrates this dynamic equilibrium. However, it's important to note that not all reactions reach complete equilibrium; instead, they reach a stage where the ratio of concentrations of reactants and products remains unchanged. Understanding this helps us predict how a system will respond to changes in temperature, pressure, or volume.
Reaction Stoichiometry
Reaction stoichiometry involves calculating the quantities of reactants and products in a chemical reaction. It uses the relationship between the balanced chemical equation and moles to determine these quantities. In the exercise provided, we're dealing with the reaction:\[\mathrm{N}_2\mathrm{O}_4 (g) \rightarrow 2\mathrm{NO}_2 (g)\]When working with stoichiometry:
  • For every mole of \( \mathrm{N}_2\mathrm{O}_4 \) decomposed, two moles of \( \mathrm{NO}_2 \) are produced.

  • In this problem, 20% of the initial \( 1 \, \text{mole} \) of \( \mathrm{N}_2\mathrm{O}_4 \) decomposes, calculating as \( 0.2 \, \text{moles} \).

  • This forms \( 0.4 \, \text{moles} \) of \( \mathrm{NO}_2 \).

Ultimately, stoichiometry allows us to determine that the total moles of gas in the container after decomposition becomes \[0.8 \, \text{moles} \; \mathrm{N}_2\mathrm{O}_4 + 0.4 \, \text{moles} \; \mathrm{NO}_2 = 1.2 \, \text{moles}.\]This information is critical in applying other gas laws and calculations.
Gas Pressure Calculation
Gas pressure in a closed container can be calculated using the Ideal Gas Law, which states that \[PV = nRT\]where:
  • \(P\) is the pressure,

  • \(V\) is the volume of the gas,

  • \(n\) is the number of moles,

  • \(R\) is the ideal gas constant, and

  • \(T\) is the temperature in Kelvin.

In this particular exercise, we use the changes in conditions (temperature and quantity of moles) to find the new pressure. The initial conditions are:
  • \(P_1 = 1 \; \text{atm}\),

  • \(n_1 = 1 \; \text{mole}\),

  • \(T_1 = 300 \; \text{K}\)

After the reaction, the conditions shift to:
  • \(n_2 = 1.2 \; \text{moles}\),

  • \(T_2 = 600 \; \text{K}\)

Since the volume remains constant, we calculate the new pressure:\[P_2 = P_1 \times \frac{n_2T_2}{n_1T_1} = 1 \times \frac{1.2 \times 600}{1 \times 300} = 2.4 \, \text{atm}\]This calculation confirms that the new pressure is 2.4 atm. Understanding how to manipulate the ideal gas law provides deeper insights into the relationships between pressure, volume, and temperature.

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Most popular questions from this chapter

A \(1 \mathrm{M}\) solution of glucose reaches dissociation equilibrium given below. $$ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} \rightleftharpoons 6 \mathrm{HCHO} $$ If the equilibrium constant is \(0.167 \times 10^{-22}\), the concentration of \(\mathrm{HCHO}\) in the equilibrium is (a) \(1.60 \times 10^{-8} \mathrm{M}\) (b) \(3.20 \times 10^{-6} \mathrm{M}\) (c) \(3.20 \times 10^{-4} \mathrm{M}\) (d) \(1.60 \times 10^{-4} \mathrm{M}\)

What is the correct sequence of active masses in increasing order in gaseous mixture, containing one gram per litre of each of the following? 1\. \(\mathrm{NH}_{3}\) 2\. \(\mathrm{N}_{2}\) 3\. \(\mathrm{H}_{2}\) 4\. \(\mathrm{O}_{2}\) Select the correct answer using the codes given below: (a) \(3,1,4,2\) (b) \(3,4,2,1\) (c) \(2,1,4,3\) (d) \(4,2,1,3\)

find \(\Delta \mathrm{G}^{\circ}\) for the reaction given below? \(\frac{1}{2} \mathrm{~A}+\frac{3}{2} \mathrm{~B} \rightleftharpoons \mathrm{C}_{\mathrm{B}}\) \(\mathrm{K}_{\mathrm{eq}}=826 \mathrm{~atm}^{-1}\) at \(298 \mathrm{~K}\) (a) \(-8.32 \mathrm{KJ}\) (b) \(8.32 \mathrm{KJ}\) (c) \(16.64 \mathrm{KJ}\) (d) \(-16.64 \mathrm{KJ}\)

A chemical reaction is catalysed by a catalyst \(X\). Hence, \(\mathrm{X}\) (a) increases activation energy of the reaction (b) does not affect equilibrium constant of the reaction (c) decreases rate constant of the reaction (d) reduces enthalpy of the reaction

If an inert gas is added in the reaction \(\mathrm{N}_{2}+3 \mathrm{H}_{2} \longrightarrow 2 \mathrm{NH}_{3}\) at constant volume, then its equilibrium (a) remains unaffected (b) favours the backward reaction (c) favours the forward reaction (d) increases the dissociation of reactants
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