The equilibrium between three substances, a, B and c is shown below.
A(g) + B(g) C(g)
Initially there were 0.1 mol of a and 0.2 mol of B in the reaction mixture. a and B reacted together to produce an
equilibrium mixture containing 0.04 mol of c. The total volume of the mixture was 2.00 dm3.
a Calculate the number of moles of a and B at equilibrium.
b Calculate the concentrations of a, B and c at equilibrium.
c i Write the equilibrium expression for Kc
.
ii Calculate the value of Kc
and give the units.
a) To find the number of moles of A and B at equilibrium, we need to use the stoichiometry of the reaction. From the reaction equation, we can see that 1 mole of A reacts with 1 mole of B to produce 1 mole of C.
Since the reaction produces 0.04 mol of C, it means that 0.04 mol of A and 0.04 mol of B have reacted. Therefore, at equilibrium, the number of moles of A and B would be:
A: 0.1 mol (initial) - 0.04 mol (reacted) = 0.06 mol
B: 0.2 mol (initial) - 0.04 mol (reacted) = 0.16 mol
So, at equilibrium, there are 0.06 moles of A and 0.16 moles of B.
b) To calculate the concentrations of A, B, and C at equilibrium, we need to divide the number of moles by the total volume of the mixture.
Concentration of A at equilibrium = 0.06 mol / 2.00 dm3 = 0.03 mol/dm3
Concentration of B at equilibrium = 0.16 mol / 2.00 dm3 = 0.08 mol/dm3
Concentration of C at equilibrium = 0.04 mol / 2.00 dm3 = 0.02 mol/dm3
Therefore, the concentrations of A, B, and C at equilibrium are 0.03 mol/dm3, 0.08 mol/dm3, and 0.02 mol/dm3, respectively.
c) i) The equilibrium expression for Kc can be written as:
Kc = [C] / ([A] * [B])
Where [C], [A], [B] are the molar concentrations of C, A, and B, respectively.
ii) To calculate the value of Kc, we substitute the given values:
Kc = (0.02 mol/dm3) / ((0.03 mol/dm3) * (0.08 mol/dm3))
Calculating the value, we get:
Kc = 8.33 dm3/mol2
The units of Kc are dm3/mol2.
a) To calculate the number of moles of a and B at equilibrium, we need to use the stoichiometry of the reaction. From the balanced equation, we can see that the mole ratio between A and C is 1:1, and between B and C is also 1:1.
Given:
Initial moles of A (a): 0.1 mol
Initial moles of B: 0.2 mol
Equilibrium moles of C: 0.04 mol
Since the mole ratio between A and C is 1:1, and the amount of C at equilibrium is 0.04 mol, the moles of A at equilibrium will also be 0.04 mol.
Similarly, since the mole ratio between B and C is 1:1, the moles of B at equilibrium will also be 0.04 mol.
Therefore, the number of moles of a and B at equilibrium is 0.04 mol each.
b) To calculate the concentrations of a, B, and C at equilibrium, we need to divide the number of moles by the total volume of the mixture.
Concentration of a = Moles of a / Total volume
= 0.04 mol / 2.00 dm^3
= 0.02 mol/dm^3
Concentration of B = Moles of B / Total volume
= 0.04 mol / 2.00 dm^3
= 0.02 mol/dm^3
Concentration of C = Moles of C / Total volume
= 0.04 mol / 2.00 dm^3
= 0.02 mol/dm^3
Therefore, the concentrations of a, B, and C at equilibrium are 0.02 mol/dm^3 each.
c) i) The equilibrium expression for Kc can be written using the concentrations of the reactants and products in the equation. For the given reaction:
A(g) + B(g) ⇌ C(g)
The equilibrium expression is:
Kc = [C] / ([A] * [B])
ii) To calculate the value of Kc, we need to substitute the equilibrium concentrations of a, B, and C into the equilibrium expression and solve for Kc.
Kc = [C] / ([A] * [B])
= (0.02 mol/dm^3) / ((0.02 mol/dm^3) * (0.02 mol/dm^3))
= 1 / 0.02
= 50
The value of Kc is 50, and the units of Kc are mol/dm^3.
a) To calculate the number of moles of a and B at equilibrium, we need to consider the balanced chemical equation and the stoichiometry of the reaction. From the equation A(g) + B(g) ⟶ C(g), we can see that the ratio of moles of a to B is 1:1.
Given that the initial moles of a is 0.1 and the initial moles of B is 0.2, and considering the stoichiometry, we can calculate the moles of a and B at equilibrium.
Since the ratio of a to B is 1:1, and we know that 0.04 mol of c is produced, the moles of a and B at equilibrium will also be 0.04 mol each.
b) To calculate the concentrations of a, B, and c at equilibrium, we need to divide the moles of each substance by the total volume of the mixture.
The total volume of the mixture is given as 2.00 dm³.
Concentration of a at equilibrium = moles of a / total volume
Concentration of a at equilibrium = 0.04 mol / 2.00 dm³
Concentration of a at equilibrium = 0.02 mol/dm³
Concentration of B at equilibrium = moles of B / total volume
Concentration of B at equilibrium = 0.04 mol / 2.00 dm³
Concentration of B at equilibrium = 0.02 mol/dm³
Concentration of c at equilibrium = moles of c / total volume
Concentration of c at equilibrium = 0.04 mol / 2.00 dm³
Concentration of c at equilibrium = 0.02 mol/dm³
c)
i) The equilibrium expression for Kc can be written based on the balanced chemical equation:
Kc = [C] / ([A] * [B])
where [A], [B], and [C] are the concentrations of substances A, B, and C, respectively, at equilibrium.
ii) To calculate the value of Kc, we need to substitute the equilibrium concentrations of A, B, and C into the equilibrium expression.
Given that the concentrations of A, B, and C at equilibrium are 0.02 mol/dm³, the value of Kc can be calculated as:
Kc = [C] / ([A] * [B])
Kc = (0.02 mol/dm³) / (0.02 mol/dm³ * 0.02 mol/dm³)
Kc = 1 / 0.02 mol²/dm⁶
Kc = 50 mol⁻²/dm²
The units of Kc are mol⁻²/dm².
a. Do this the same way as the previous problem for moles.
b. mols/volume in dm^3 = concentration.
c. Kc = (C)/(A)(B)
ii. Plug in the concentration values from part b and calculate Kc.
Post your work if you get stuck.