Iodine is an essential trace element for life and is the heaviest element commonly needed by living — Analytical Chemistry Chemistry Question
Chemical Equilibrium
Iodine is an essential trace element for life and is the heaviest element commonly needed by living organisms. At high temperatures an equilibrium between I2(g) and I(g) takes place.
The following table summarizes the initial pressure of I2(g) and the total pressure when the equilibrium is reached at the given temperatures.
[VISUAL]
Table data:
T (K): 1073, 1173
p(I2) (atm): 0.0631, 0.0684
ptotal (atm): 0.0750, 0.0918
Calculate ∆H°, ∆G° and ∆S° at 1100 K. (Assume that ∆H° and ∆S° are independent on temperature in the temperature range given.)
Model Answer
At equilibrium p(I2)eq = p(I2)0 – x. Thus, ptotal = p(I2)0 + x.
At 1073 K, x = 0.0750 – 0.0631 = 0.0119 bar
p(I)eq = 2x = 0.0238 bar
p(I2)eq = 0.0631 – 0.0119 = 0.0512 bar
Kp = (p(I)eq)^2 / p(I2)eq = 0.0238^2 / 0.0512 = 0.0111
At 1173 K, x = 0.0918 – 0.0684 = 0.0234 bar
p(I)eq = 2x = 0.0468 bar
p(I2)eq = 0.0684 – 0.0234 = 0.0450 bar
Kp = (p(I)eq)^2 / p(I2)eq = 0.0468^2 / 0.0450 = 0.0487
Using Clausius-Clapeyron/Van 't Hoff equation:
ln(Kp2 / Kp1) = -Delta H°/R * (1/T2 - 1/T1)
ln(0.04867 / 0.01106) = 1.4817
(1/1073 - 1/1173) = 7.945 * 10^-5 K^-1
Delta H° = 1.4817 * 8.314 / (7.945 * 10^-5) = 155 kJ
At 1100 K:
ln(K1100 / 0.01106) = (155052 / 8.314) * (1/1073 - 1/1100)
K1100 = 0.017
Delta G° = -RT ln K = -8.314 * 1100 * ln(0.017) = 37.3 kJ/mol (37263 J/mol)
Delta S° = (Delta H° - Delta G°) / T = (155052 - 37263) / 1100 = 107 J K^-1 mol^-1
Calculate the mole fraction of I(g) in the equilibrium mixture when the numerical value of Kp is the half of the total pressure.
Model Answer
I2(g) <=> 2 I(g)
p(I2)e = p(I2)0 – x
ptotal = p(I2)0 + x
Kp = 4x^2 / (p(I2)0 - x) = ptotal / 2 = (p(I2)0 + x) / 2
8x^2 = (p(I2)0 + x)(p(I2)0 - x) = p(I2)0^2 - x^2
9x^2 = p(I2)0^2 ⇒ p(I2)0 = 3x
ptotal = p(I2)0 + x = 4x
p(I)eq = 2x
Mole fraction of I(g) = p(I)eq / ptotal = 2x / 4x = 0.50
Assuming ideal gas behavior for I2(g) and I(g), calculate the bond energy of I2 at 298 K.
Model Answer
For an ideal monatomic gas Cp,m = Cv,m + R = 3/2 R + R = 5/2 R
For an ideal diatomic gas Cp,m = Cv,m + R = 5/2 R + R = 7/2 R
Delta Cp = 2 * Cp,m(I) - Cp,m(I2) = 2 * (2.5 R) - 3.5 R = 1.5 R
Delta H°298 = bond energy I-I
Delta T = 298 - 1100 = -802 K
Delta H°298 = Delta H°1100 + Delta Cp * Delta T
= 155052 + (298 - 1100) * (2 * 2.5 - 3.5) * R = 155052 - 10001 = 145 kJ
Calculate the wavelength of radiation that must be used to dissociate I2(g) at 298 K.
Model Answer
Delta H°298 = E = N_A * h * c / lambda
lambda = N_A * h * c / E
= (6.022 * 10^23 * 6.63 * 10^-34 * 3.00 * 10^8) / 145050 = 825.8 nm = 826 nm
In an experiment, when a sample of I2(g) is irradiated by a laser beam of λ = 825.8 nm, at a rate of 20.0 J s-1 for 10.0 s, 1.0 ⋅ 10-3 mol of I(g) is produced. Calculate the quantum yield for the dissociation process (i.e., the number of moles of I2 dissociated per mole of photons absorbed by the system).
Model Answer
Ephoton = h * c / lambda = (6.63 * 10^-34 * 3.00 * 10^8) / (825.8 * 10^-9) = 2.409 * 10^-19 J
Total energy absorbed = P * t = 20.0 J s^-1 * 10.0 s = 200 J
Nphotons = 200 / (2.409 * 10^-19) = 8.30 * 10^20
n_photons = Nphotons / N_A = 8.30 * 10^20 / (6.02 * 10^23) = 1.38 * 10^-3 mol photons
n_I2_dissoc = 0.5 * n_I = 0.5 * 1.0 * 10^-3 mol = 5.0 * 10^-4 mol
Quantum yield = 5.0 * 10^-4 / (1.38 * 10^-3) = 0.36