at 22°C the equilibrium constant, Kc, for the following reaction is 4.66 x 10-3. (1) If 0.800 mol of N2O4 were injected into a closed 1.00-L container at 22°C, how many moles of each gas would be present at equilibrium? (b) If the volume were halved (to 0.500L) at constant temperature, how many moles of each gas would be present after the new equilibrium has been established? N2O4(g) 2NO2(g)
To solve this problem, we will use the given equilibrium constant (Kc) and the stoichiometry of the balanced chemical equation.
The balanced chemical equation for the reaction is:
N2O4(g) ↔ 2NO2(g)
(a) Initial conditions:
- Initial moles of N2O4 = 0.800 mol
- Initial moles of NO2 = 0 mol
Let x moles of N2O4 react to reach equilibrium. This causes the formation of 2x moles of NO2.
The equilibrium concentrations can be calculated using the equation:
Kc = ([NO2]^2) / [N2O4]
Substituting the given values:
4.66 x 10^-3 = (2x)^2 / (0.800 - x)
Since the volume is constant, the total moles of the system remain constant. Therefore, we can assume that at equilibrium, the moles of N2O4 reacted (x) will equal the moles of NO2 formed (2x). Using this relationship, we can simplify the equation:
4.66 x 10^-3 = (2x)^2 / (0.800 - x)
4.66 x 10^-3 * (0.800 - x) = 4x^2
Solving this equation will give us the value of x, which will correspond to the moles of each gas at equilibrium.
(b) If the volume is halved (0.500 L), the concentration of each gas will change, but the equilibrium constant remains the same.
Using the moles of N2O4 reacted (x) calculated in part (a), we can calculate the moles of each gas at equilibrium in the new volume (0.500 L) using the stoichiometry:
Moles of N2O4 at equilibrium: 0.800 - x
Moles of NO2 at equilibrium: 2x
Please note that to obtain the exact numerical values, you need to solve the equation in part (a) for x.
To solve this question, we will use the equilibrium constant expression and the given information.
The balanced equation for the reaction is:
N2O4(g) ↔ 2NO2(g)
(a) To find the number of moles of each gas at equilibrium when 0.800 mol of N2O4 is injected into a 1.00-L container, we need to use the equilibrium constant expression:
Kc = [NO2]^2 / [N2O4]
Given that Kc = 4.66 x 10^-3 and [N2O4] = 0.800 mol, we can rearrange the equation to solve for [NO2]:
4.66 x 10^-3 = [NO2]^2 / 0.800
Multiply both sides by 0.800 to isolate [NO2]^2:
(4.66 x 10^-3) * 0.800 = [NO2]^2
Solving this equation gives us:
[NO2]^2 = 3.728 x 10^-3
Taking the square root of both sides, we get:
[NO2] ≈ 0.061 mol
Since the reaction is N2O4(g) ↔ 2NO2(g), we know that for every one mole of N2O4, two moles of NO2 are produced. Therefore, the number of moles of N2O4 remaining at equilibrium will be:
[N2O4] = initial moles - moles reacted
= 0.800 - (0.061 * 2)
≈ 0.678 mol
So, at equilibrium, there will be approximately 0.061 mol of NO2 and 0.678 mol of N2O4.
(b) If the volume is halved to 0.500 L while maintaining a constant temperature, the number of moles of each gas at the new equilibrium can be calculated using the same equilibrium constant expression:
Kc = [NO2]^2 / [N2O4]
Since the temperature remains constant, the value of Kc remains the same. Let's assume the number of moles of NO2 and N2O4 at the new equilibrium are [NO2'] and [N2O4'], respectively.
Kc = ([NO2']^2) / [N2O4']
We also know that the initial concentration of N2O4 does not change, which means [N2O4] = 0.800 mol.
To find the new equilibrium concentrations, we will use the volume ratio of the new container to the initial container:
Volume ratio = final volume / initial volume
= 0.500 L / 1.00 L
= 0.500
According to the ideal gas law, the concentration is directly proportional to the volume. So, the concentration of both NO2 and N2O4 will be halved.
[N2O4'] = [N2O4] / Volume ratio
= 0.800 mol / 0.500
= 1.600 mol
Since the number of moles of NO2 is twice that of N2O4 when the reaction reaches equilibrium, we have:
[NO2'] = 2 * [N2O4']
= 2 * 1.600
= 3.200 mol
Therefore, at the new equilibrium, there will be approximately 3.200 mol of NO2 and 1.600 mol of N2O4.
..........N2O4 ==> 2NO2
I........0.800.......0
C.........-x........2x
E........0.800-x....2x
Substitute the E line into Kc and calculate x and evaluate 0.800 -x. This give you M, the volume is 1 L, solve for mols.
b)
Calculate new (N2O4)= 0.800/0.5 = 1.600
Redo part a as above, x will be M, determine mols in 0.5L.