Suppose that A and B react to form C according to the equation below.
A + 2 B ⇌ C
What are the equilibrium concentrations of A, B and C if 3.0 mol A and 3.0 mol B are added to a 1.0 L flask? Assume that the equilibrium constant for this reaction is Kc = 8.1x1013.
.................A + 2 B ⇌ C
I...............3M......3M.....0
C.............-x..........-2x.....x
E.............3-x.......3-2x.....x
Plug the E line into the Kc expression and solve for x, then evaluate each item in the E line to find the individual concentrations. Post your work if you get stuck.
To determine the equilibrium concentrations of A, B, and C, we need to use the equilibrium constant, Kc, and apply the equation relating concentrations to the equilibrium constant.
The equation A + 2B ⇌ C represents a chemical reaction where A and B react to form C. The stoichiometric coefficients of A and B suggest that the reaction consumes one mole of A for every two moles of B.
Given that 3.0 mol A and 3.0 mol B are added to a 1.0 L flask, we can start by assuming the initial concentration of A is [A]₀ = 3.0 M and the initial concentration of B is [B]₀ = 3.0 M. Since no C is present initially, [C]₀ = 0 M.
At equilibrium, the concentrations of A, B, and C will be denoted as [A]eq, [B]eq, and [C]eq, respectively.
The equation for the equilibrium constant, Kc, is defined as follows:
Kc = ([C]eq) / ([A]eq * ([B]eq)²)
We can rearrange this equation to solve for [C]eq:
[C]eq = Kc * [A]eq * [B]eq²
Now, let's calculate the equilibrium concentrations of A, B, and C:
1. Substitute the given values:
Kc = 8.1x10¹³
[A]eq = [A]₀ = 3.0 M
[B]eq = [B]₀ = 3.0 M
2. Plug these values into the equation and solve for [C]eq:
[C]eq = (8.1x10¹³) * (3.0 M) * (3.0 M)²
Calculating this expression gives the equilibrium concentration of C, [C]eq.
Please note that [A]eq and [B]eq will remain 3.0 M since there is no reaction occurring that generates or consumes A or B after equilibrium is reached.