- #1
ScrubsFan
- 15
- 0
Consider the following reaction
A(g) <---> 2B(g) + C(g)
When 1.00 mol of A is placed in a 4.00L container at temperature t, the concentration of C at equilibrium is 0.050 mol/L. What is the equilibrium constant for the reaction at temperature t?
Here is my answer:
The balanced equation shows that for each mol of C formed, 2 mol of B is formed and 1 mol of A is consumed. We are told that, at equilibrium, 0.050 mol of C has been produced. This means that 0.050 mol of B must also have been produced and that 0.050 mol of A was consumed. Therefore, at equilibrium,
0.050 mol C x 2mol B / 1mol C = 0.1 mol B
1.00 mol A - [0.050 mol C x 1mol A / 2mol C]
= 0.975 / 0.98 mol A
and the equilibrium concentrations are:
[C] = 0.050 mol / 4.00 L = 0.0125 mol/L
2 = 0.1 mol / 4.00 L = 0.025 mol/L
[A] = 0.98 mol / 4.00 L = 0.245 mol/L
Substitute the equilibrium concentrations in the equilibrium law expression and solve for Ke:
Ke = 2[C] / [A]
= 0.025 mol/L x 0.0125 mol/L
0.245 mol/L
= 0.0013 mol/L
A(g) <---> 2B(g) + C(g)
When 1.00 mol of A is placed in a 4.00L container at temperature t, the concentration of C at equilibrium is 0.050 mol/L. What is the equilibrium constant for the reaction at temperature t?
Here is my answer:
The balanced equation shows that for each mol of C formed, 2 mol of B is formed and 1 mol of A is consumed. We are told that, at equilibrium, 0.050 mol of C has been produced. This means that 0.050 mol of B must also have been produced and that 0.050 mol of A was consumed. Therefore, at equilibrium,
0.050 mol C x 2mol B / 1mol C = 0.1 mol B
1.00 mol A - [0.050 mol C x 1mol A / 2mol C]
= 0.975 / 0.98 mol A
and the equilibrium concentrations are:
[C] = 0.050 mol / 4.00 L = 0.0125 mol/L
2 = 0.1 mol / 4.00 L = 0.025 mol/L
[A] = 0.98 mol / 4.00 L = 0.245 mol/L
Substitute the equilibrium concentrations in the equilibrium law expression and solve for Ke:
Ke = 2[C] / [A]
= 0.025 mol/L x 0.0125 mol/L
0.245 mol/L
= 0.0013 mol/L