0.80 mol/L of A and B are mixed. They react very slowly to produce C and D as follows:
A(aq) + B(aq) ⇔ C(aq) + 2D(aq)
When equilibrium is attained, the concentration of C is 0.60 mol/L and the concentration of A is:
a) 0.20 M
b) 0.40 M
c) 0.60 M
d) 0.80 M
.......A + B ==> C + D
I.....0.8..0.8...0...0
C.....-x...-x....x...x
E...0.8-x.0.8-x..x...x
So C = x = 0.6. and
0.8-x = A = ?
0.2
To determine the concentration of A at equilibrium, we can use the stoichiometry of the balanced chemical equation and the concentration of C in order to set up an equilibrium expression.
The balanced chemical equation is:
A(aq) + B(aq) ⇔ C(aq) + 2D(aq)
The equilibrium expression is given by:
K = [C][D]^2 / [A][B]
Since the reaction is at equilibrium, we can assume that the concentrations of C and D do not change anymore. We can use the given concentration of C, which is 0.60 mol/L, and substitute it into the equilibrium expression.
K = (0.60)[D]^2 / (concentration of A)[B]
Now, we are given that the initial concentrations of A and B are both 0.80 mol/L. At equilibrium, we can assume that the concentration of B does not change, so we can substitute 0.80 into the expression.
K = (0.60)[D]^2 / (concentration of A)(0.80)
Next, we need to determine the value of K. This value depends on the specific reaction conditions, such as temperature and pressure. Without this information, we cannot calculate the exact value of K. However, we can determine the ratio of the concentrations of products to reactants at equilibrium, which will allow us to compare the concentrations.
If K is greater than 1, it means that the concentration of products is higher at equilibrium, indicating that the forward reaction is favored. Conversely, if K is less than 1, it means that the concentration of reactants is higher at equilibrium, indicating that the reverse reaction is favored.
Since we don't have the value of K, we cannot determine whether the forward or reverse reaction is favored. However, we can make an assumption based on the given choices for the concentration of A.
Among the given choices (a) 0.20 M, (b) 0.40 M, (c) 0.60 M, and (d) 0.80 M, the closest concentration to the initial concentration of A (0.80 M) is option (d) 0.80 M.
Therefore, the concentration of A at equilibrium is approximately 0.80 M.