Lab 6 (R)

Question 1

Objectives

Answer 1

To determine the values of the thermodynamic parameters ∆H°, ∆G°, and ∆S° for the reaction:

Fe3+(aq) + SCN–(aq) ⇌ [Fe(SCN)]2+(aq)

Question 2

The equilibrium constant, Keq, for the reaction

Fe3+(aq) + SCN–(aq) ⇌ [Fe(SCN)]2+(aq)

Answer 2

K_eq = [[Fe(SCN)]²⁺]_eq / [Fe³⁺]_eq -[SCN^-]_eq

Question 3

Value for the equilibrium constant can easily be determined for this reaction (_{Fe3+(aq) + SCN–(aq) ⇌ [Fe(SCN)]2+(aq)})

Answer 3

since the complex ion [Fe(SCN)]2+ absorbs visible light at 447 nm.

Spectrophotometer is used to determine the concentration of [[Fe(SCN)]2+]eq.

Knowing the initial reagent concentration, the concentration of all entities at equilibrium can be determined using an ICE table.

When these concentrations are substituted into the expression for Keq, the equilibrium constant can be determined.

Question 4

The equilibrium constant is constant at ____, but is dependent on ____.

It is this dependence on ____ that can be used to extract the thermodynamic parameters ___, ___, and ___.

Answer 4

1. A fixed temperature
2. Temperature
3. Temperature
4. ∆H°
5. ∆G°
6. ∆S°

Question 5

For any reaction, there exists a relationship between standard free energy (∆G°) and the equilibrium constant (Keq):

equation _____.

When this is combined with the defining equation for free energy, ____. (equation)

the following relationships are derived:
_____, _____, _____.

Answer 5

1. ∆G° = –RT ln(Keq) (

where R is the gas constant (8.3145 J K–1 mol–1) and T is the temp in Kelvin.

2. ∆G° = ∆H°–T∆S°
3. –RT ln(Keq) = ∆G° = ∆H°–T∆S°
4. ln(Keq) = - ∆H° / RT + ∆S° / R
   _{(by dividing both sides of (2) by -RT)}
5. ln(Keq) = ∆H° / RT (1/T) + ∆S° / R
   _{(simply (4) expressed slightly differently)}

Question 6

If we assume that ∆H° and ∆S° are not highly temperature dependent (a reasonable assumption), then ____ should give a straight line with: ____, ____.

The slope and y-intercept could be ________, depending on the value of _____.

Answer 6

1. a plot of ln(Keq) vs. 1/T
2. slope = - ∆H° / R
3. y-intercept = ∆S° / R
4. positive or negative
5. ∆H° and ∆S°

Question 7

_{Advance Study Assignment}

1. Beer’s Law relates absorbance to concentration, path length and molar absorptivity: A = ε·b·c (b is the cuvette path length in cm and ε is the molecule’s molar absorptivity, at a given wavelength, with units (mol/L)–1 cm–1).

In this experiment, b = 1.00 cm and ε = 2.20·103 L/(mol cm) for Fe(SCN)2+ at 447 nm.

_{(This data is hypothetical and is in no way meant to convey expected experimental results – these data are provided for instructional purposes only).}

If [Fe3+]stock = 0.00123 mol/L, [SCN–]stock = 0.00234 mol/L and 5.00 mL of Fe3+(aq) was combined with 3.00 mL of SCN–(aq) at 12 °C, the resulting equilibrium mixture has an absorbance of 0.100 at 12 °C. The same equilibrium mixture had an absorbance of 0.200 at
84 °C.

a) Calculate Keq at 12 °C and at 84 °C.

Answer 7

a) Dilution must be taken into account.

[Fe³⁺]_diluted = [Fe³⁺]_stock × V_Fe³⁺ / V_total = 0.00123 mol/L × (5.00 mL/8.00 mL) = 0.0007688 mol/L

[SCN^-]_diluted = [SCN^-]_stock × V_SCN^- / V_total = 0.00234 mol/L × (3.00 mL/8.00 mL) = 0.0008775 mol/L

At 12 °C

[[Fe(SCN)]²⁺] = A/(ε ·b) = 0.100 ÷ (1.00 cm · 2.20·10³ L/(mol cm)) = 0.00004545 mol/L

Since the reaction (1) is a 1: 1: 1 ratio, [[Fe(SCN)]²⁺] = [Fe³⁺]_Reacted = [SCN^-]_Reacted

[Fe³⁺]_eq = [Fe³⁺]_diluted - [Fe³⁺]_Reacted

[SCN^-]_eq = [SCN^-]_diluted - [SCN^-]_Reacted

[Fe³⁺]_eq = 0.0007688 mol/L ⎯ 0.00004545 mol/L = 0.0007234 mol/L

[SCN^-]_eq = 0.0008775 mol/L ⎯ 0.00004545 mol/L = 0.0008321 mol/L

k_eq = [[Fe(SCN)]²⁺]_eq / [Fe³⁺]_eq [SCN^-]_eq = 0.00004545 / (0.0007234)(0.0008321) = 75.4 (at 12°C)

At 84 °C

[[Fe(SCN)]²⁺] = A/(ε ·b) = 0.200 ÷ (1.00 cm · 2.20·10³ L/(mol cm)) = 0.0000909 mol/L

[Fe³⁺]_eq = = 0.0007688 mol/L ⎯ 0.0000909 mol/L = 0.0006779 mol/L

[SCN^-]_eq = 0.0008775 mol/L ⎯ 0.0000909 mol/L = 0.0007866 mol/L

k_eq = [[Fe(SCN)]²⁺]_eq / [Fe³⁺]_eq [SCN^-]_eq = 0.0000909 / (0.0006779)(0.0007866) = 170.5 (at 84°C)

Question 8

1. In this experiment, b = 1.00 cm and ε = 2.20·103 L/(mol cm) for Fe(SCN)2+ at 447 nm.

If [Fe3+]stock = 0.00123 mol/L, [SCN–]stock = 0.00234 mol/L and 5.00 mL of Fe3+(aq) was combined with 3.00 mL of SCN–(aq) at 12 °C, the resulting equilibrium mixture has an absorbance of 0.100 at 12 °C. The same equilibrium mixture had an absorbance of 0.200 at
84 °C.

b) Calculate ∆H° for the reaction.

Answer 8

b) A series of data is graphed ln K vs. 1/T (see page R-2).

The linear equation (of the form y = mx +b) for the best-fit line for the data plotted above is y = -1153K (1/T) + 8.36.

The slope of the line of best fit is -1153 K.

Slope= –(∆H°/R).
∆H° = –mR
∆H° = –(–1153 K ∙ 8.3145 JK-1mol-1)
∆H° = 9587 J mol-1

Question 9

1. In this experiment, b = 1.00 cm and ε = 2.20·103 L/(mol cm) for Fe(SCN)2+ at 447 nm.

If [Fe3+]stock = 0.00123 mol/L, [SCN–]stock = 0.00234 mol/L and 5.00 mL of Fe3+(aq) was combined with 3.00 mL of SCN–(aq) at 12 °C, the resulting equilibrium mixture has an absorbance of 0.100 at 12 °C. The same equilibrium mixture had an absorbance of 0.200 at
84 °C.

c) Calculate ∆S° for the reaction.

Answer 9

c) The y-intercept of the line of best fit is 8.36.

y-intercept = (∆S°/R).
ΔS° = 8.36 ∙ 8.3145 JK-1mol-1
ΔS° = 69.5 J mol-1 K-1

Question 10

1. In this experiment, b = 1.00 cm and ε = 2.20·103 L/(mol cm) for Fe(SCN)2+ at 447 nm.

If [Fe3+]stock = 0.00123 mol/L, [SCN–]stock = 0.00234 mol/L and 5.00 mL of Fe3+(aq) was combined with 3.00 mL of SCN–(aq) at 12 °C, the resulting equilibrium mixture has an absorbance of 0.100 at 12 °C. The same equilibrium mixture had an absorbance of 0.200 at
84 °C.

d) Calculate ∆G° for the reaction at 12 °C and 84 °C.

Answer 10

d) ∆G° at 12 °C may be calculated using either equation (1) or (2), both yield the same result.

Ex. (1) will be used to calculate ∆G° at 12 °C.

∆G° at 12 °C: ∆G° = –RT ln(Keq) = –8.3145 J K–1 mol–1(285 K) × ln(75.4) = –10.2 kJ mol–1

Ex. (2) will be used to calculate ∆G° at 84 °C.

∆G° at 84 °C: ∆G° = ∆H°– T∆S° = 9587 J mol–1 – 357 K(69.6 J K–1 mol–1) = –15.3 kJ mol–1