When 0.50 Mol Of $N_2O_4$ Is Placed In A 4.0-liter Reaction Vessel And Heated To 400 K, $80\%$ Of The $N_2O_4$ Decomposes To $NO_2$ Gas As Follows:$\[ N_2O_4(g) \rightleftharpoons 4NO(g) \\]What Will Be The

When 0.50 mol of $N_2O_4$ is placed in a 4.0-liter reaction vessel and heated to 400 K

In this article, we will explore the decomposition of $N_2O_4$ to $NO_2$ gas in a 4.0-liter reaction vessel at a temperature of 400 K. The reaction is as follows:

${ N_2O_4(g) \rightleftharpoons 2NO_2(g) }$

Understanding the Reaction

The reaction is a reversible reaction, meaning that it can proceed in both forward and backward directions. The forward reaction involves the decomposition of $N_2O_4$ to form $NO_2$ gas, while the backward reaction involves the recombination of $NO_2$ gas to form $N_2O_4$. The equilibrium constant for the reaction is given by:

${ K_c = \frac{[NO_2]^2}{[N_2O_4]} }$

Calculating the Initial Concentration of $N_2O_4$

The initial concentration of $N_2O_4$ can be calculated using the ideal gas law:

${ PV = nRT }$

where $P$ is the pressure, $V$ is the volume, $n$ is the number of moles, $R$ is the gas constant, and $T$ is the temperature.

Rearranging the equation to solve for $n$, we get:

${ n = \frac{PV}{RT} }$

Substituting the given values, we get:

${ n = \frac{(1 \text{ atm})(4.0 \text{ L})}{(0.0821 \text{ L atm/mol K})(400 \text{ K})} = 0.0495 \text{ mol} }$

However, we are given that 0.50 mol of $N_2O_4$ is placed in the reaction vessel. Therefore, the initial concentration of $N_2O_4$ is:

${ [N_2O_4]_0 = \frac{0.50 \text{ mol}}{4.0 \text{ L}} = 0.125 \text{ M} }$

Calculating the Concentration of $NO_2$

The reaction is a reversible reaction, and 80% of the $N_2O_4$ decomposes to form $NO_2$ gas. Therefore, the concentration of $NO_2$ can be calculated as follows:

${ [NO_2] = 0.8[N_2O_4]_0 = 0.8(0.125 \text{ M}) = 0.1 \text{ M} }$

Calculating the Equilibrium Constant

The equilibrium constant for the reaction can be calculated using the following equation:

${ K_c = \frac{[NO_2]^2}{[N_2O_4]} }$

Substituting the values, we get:

${ K_c = \frac{(0.1 \text{ M})^2}{(0.125 \text{ M})} = 0.08 \text{ M} }$

In this article, we have explored the decomposition of $N_2O_4$ to $NO_2$ gas in a 4.0-liter reaction vessel at a temperature of 400 K. We have calculated the initial concentration of $N_2O_4$, the concentration of $NO_2$, and the equilibrium constant for the reaction. The results show that 80% of the $N_2O_4$ decomposes to form $NO_2$ gas, and the equilibrium constant for the reaction is 0.08 M.

• Atkins, P. W. (1998). Physical Chemistry. Oxford University Press.
• Chang, R. (2008). Chemistry. McGraw-Hill.
• Levine, I. N. (2009). Physical Chemistry. McGraw-Hill.

• Equilibrium constant: A measure of the ratio of the concentrations of the products to the concentrations of the reactants at equilibrium.
• Ideal gas law: A mathematical equation that describes the behavior of an ideal gas.
• Reversible reaction: A reaction that can proceed in both forward and backward directions.
• Equilibrium: A state in which the concentrations of the reactants and products remain constant over time.
  Q&A: When 0.50 mol of $N_2O_4$ is placed in a 4.0-liter reaction vessel and heated to 400 K

Q: What is the initial concentration of $N_2O_4$ in the reaction vessel?

A: The initial concentration of $N_2O_4$ can be calculated using the ideal gas law. Rearranging the equation to solve for $n$, we get:

${ n = \frac{PV}{RT} }$

Substituting the given values, we get:

${ n = \frac{(1 \text{ atm})(4.0 \text{ L})}{(0.0821 \text{ L atm/mol K})(400 \text{ K})} = 0.0495 \text{ mol} }$

However, we are given that 0.50 mol of $N_2O_4$ is placed in the reaction vessel. Therefore, the initial concentration of $N_2O_4$ is:

${ [N_2O_4]_0 = \frac{0.50 \text{ mol}}{4.0 \text{ L}} = 0.125 \text{ M} }$

Q: How much of the $N_2O_4$ decomposes to form $NO_2$ gas?

A: The reaction is a reversible reaction, and 80% of the $N_2O_4$ decomposes to form $NO_2$ gas. Therefore, the concentration of $NO_2$ can be calculated as follows:

${ [NO_2] = 0.8[N_2O_4]_0 = 0.8(0.125 \text{ M}) = 0.1 \text{ M} }$

Q: What is the equilibrium constant for the reaction?

A: The equilibrium constant for the reaction can be calculated using the following equation:

${ K_c = \frac{[NO_2]^2}{[N_2O_4]} }$

Substituting the values, we get:

${ K_c = \frac{(0.1 \text{ M})^2}{(0.125 \text{ M})} = 0.08 \text{ M} }$

Q: What is the significance of the equilibrium constant?

A: The equilibrium constant is a measure of the ratio of the concentrations of the products to the concentrations of the reactants at equilibrium. It is a useful tool for predicting the direction of a reaction and the extent of a reaction.

Q: Can the reaction be reversed?

A: Yes, the reaction can be reversed. The reaction is a reversible reaction, meaning that it can proceed in both forward and backward directions.

Q: What is the temperature of the reaction?

A: The temperature of the reaction is 400 K.

Q: What is the volume of the reaction vessel?

A: The volume of the reaction vessel is 4.0 liters.

Q: What is the pressure of the reaction vessel?

A: The pressure of the reaction vessel is 1 atm.

In this article, we have answered some of the frequently asked questions about the decomposition of $N_2O_4$ to $NO_2$ gas in a 4.0-liter reaction vessel at a temperature of 400 K. We have calculated the initial concentration of $N_2O_4$, the concentration of $NO_2$, and the equilibrium constant for the reaction. The results show that 80% of the $N_2O_4$ decomposes to form $NO_2$ gas, and the equilibrium constant for the reaction is 0.08 M.

• Atkins, P. W. (1998). Physical Chemistry. Oxford University Press.
• Chang, R. (2008). Chemistry. McGraw-Hill.
• Levine, I. N. (2009). Physical Chemistry. McGraw-Hill.

• Equilibrium constant: A measure of the ratio of the concentrations of the products to the concentrations of the reactants at equilibrium.
• Ideal gas law: A mathematical equation that describes the behavior of an ideal gas.
• Reversible reaction: A reaction that can proceed in both forward and backward directions.
• Equilibrium: A state in which the concentrations of the reactants and products remain constant over time.