Question

Can someone help me with 2 and 3? I don't know the answer and I need justifications for why the correct answers are correct

Answer 1

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