In lab, I added 3 (g) "solid" Urea CO(NH2)2 to a beaker and filled a buret with 50 ml H2O. I attached a thermometer
to the beaker. I added 1 ml of H2O from buret until "solid" Urea dissolved. I had to record the temperature with each
1 ml added.
I had to repeat this experiment 2 more times using different amounts of "solid" Urea each time.
The first time: I used 3(g) Urea beginning temp = 21.00 degrees C.
After first 1ml H2O temp started decreasing. The lowest temp recorded was 20.50 degrees C. until Urea dissolved.
The second time: I used 4(g) Urea beginning temp = 21.00 degrees C.
After first 1ml of H2O temperature started decreasing again. The lowest temp recorded was 20.53 degrees C.
The third time: I used 5(g) Urea beginning temp = 21.00 degrees C.
After first 1ml of H20, temperature started decreasing then it increased. The temperature dropped to 20.34 degrees C
and then rose to 20.82 degrees C.
Questions:
1. Calculate the equilibrium constant Keq by dividing the number of moles of urea in the system by total volume of
water added to the beaker in L.
First calculate moles of Urea. (Molar mass of Urea = 60.07 g/mol)
Trial 1) Moles Urea = (3 g / 60.07 g/mol )= 0.0499 mole urea
Volume of water needed to dissolve = 3 mL = 0.003 L
Keq = 0.0499 mole urea / 0.003 L = 16.63
Trial 2) Moles Urea = (4 g / 60.07 g/mol )= 0.0666 mole urea
Volume of water needed to dissolve = 4 mL = 0.004 L
Keq = 0.0666 mole urea / 0.004 L = 16.65
Trial 3) Moles Urea = (5 g / 60.07 g/mol )= 0.0832 mole urea
Volume of water needed to dissolve = 5 mL = 0.005 L
Keq = 0.0832 mole urea / 0.005 L = 16.64
2. Calculation of Delta G and Delta S.
3. Calculate the free energy change of dissolution reaction using the following formula: Delta G= =RT In Keq where R
is the gas constant (8.31451 J K-1 mol-l) and T is the temperature in Kelvin.
Trial 1
Delta G = -RT Ln(Keq) = -(8.314)(293.65K)Ln(16.63) = -6863 J
Trial 2
Delta G = -RT Ln(Keq) = -(8.314)(293.68K)Ln(16.65) = -6867 J
Trial 3
Delta G = -RT Ln(Keq) = -(8.314)(293.97K)Ln(16.64) = -6872 J
4. Using the values of Delta H, Delta G, and the temperature of the saturated solution, calculate Delta S using Gibbs
Free Energy.
The following equation related G, H, and S
G = H – T*S
Solve for S:
S = (H – G) / T
Trial 1
Delta S = (H – G) / T = (-251.54 J – 6863 J) / 293.65 = -24.23 J/K
Trial 2
Delta S = (H – G) / T = (-236.22 J – 6867 J) / 293.68 = -24.19 J/K
Trial 3
Delta S = (H – G) / T = (-90.52 J – 6872 J) / 293.97 = -23.68 J/K
Observations:
Volume of
Mass of Urea
water
Initial
Final
dissolved, m
required for
temperature
temperature
(g)
dissolution,
(T1 deg C)
(T2 deg C)
V (mL)
Change in
temperature =
T = (T2 – T1)
deg C
Trial 1
3
3
21
20.50
-0.50
Trial 2
4
4
21
20.53
-0.47
Trial 3
5
5
21
20.82
-0.18
Calculations:
Molar mass of Urea CO(NH2)2:
1 C = 12g/mol ( 20 % of mass)
1 O = 16 g/mol
( 27 % of mass)
2 N = 28 g/mol ( 47 % of mass)
4 H = 4 g/mol ( 6 % of mass)
Total:
60.07 g/mol
(1) Heat of dissolution of urea, Q = (m + M) * s * T
Assuming that 1 mL of water has a mass of 1 gram, and taking s = 4.184 J/g deg C
Trail 1: Q = (3+3) *4.184 J/g deg C *-0.50 C = -12.552 J
Trial 2: Q = (4+4) *4.184 J/g deg C*-0.47 C = -15.732 J
Trial 3: Q = (5+5) *4.184 J/g deg C*-0.18 C = -7.531 J
Average heat of dissolution of urea in water, Q = 11.938 J
(2) Molar enthalpy change, H = Q/moles of the solute dissolved
Get moles of solute from question #1
Trail 1: H =-12.552 J/(0.0499 mole) = -251.54 J/mole
Trial 2: H = -15.732 J /(0.0666 mole) = -236.22 J/mole
Trial 3: H = -7.531 J /(0.0832 mole) = -90.52 J/mole
Average molar enthalpy change, H = -192.76 J/mole.