Answer:
2. (C) K < 1.
3. (B) [Fe³⁺] = 2.00 mol·L⁻¹; [SCN⁻] = 6.0 mol·L⁻¹
Step-by-step explanation:
2. Value of K
A⇌ B
K = [B]/[A]
If the concentration of reactants (A) is larger than the concentration of products (B), the denominator of the K expression is larger than the numerator.
The fraction is less than 1, so
K < 1
3. Equilibrium concentrations
We can use an ICE table to keep track of the calculations.
Fe³⁺ + SCN⁻ ⇌ FeSCN²⁺
I/mol·L⁻¹: 6.00 10.0 0
C/mol·L⁻¹: -x -x +x
E/mol·L⁻¹: 6.00-x 10.0-x x
Initially, there is no FeSCN²⁺ present, so [FeSCN²⁺] = 0.
Then, the Fe³⁺ and SCN⁻ react until equilibrium is reached.
How much will react? We don't know, but we have every confidence that x mol (some unknown quantity) will react.
[Fe³⁺] will decrease by x mol·L⁻¹. Because of the 1:1:1 molar ratios, [SCN⁻] will also decrease by x mol·L⁻¹ and [FeSCN²⁺] will increase by x mol·L⁻¹.
We add the changes and get the values in the bottom line.
However, what is the value of that pesky x?
We are told that [FeSCN²⁺] = 4.00 mol·L⁻¹ at equilibrium.
From the table, x = [FeSCN²⁺], so x = 4.00.
Now we can insert these values back into the table.
At equilibrium,
[Fe³⁺] = 6.00 - x = 6.00 - 4.00 = 2.00 mol·L⁻¹
[SCN⁻] = 10.0 - x = 10.0 - 4.00 = 6.0 mol·L⁻¹