Answer Questions 2 and 3 1. 20 points Given the reactions () and (2) below, dete
ID: 1030479 • Letter: A
Question
Answer Questions 2 and 3
1. 20 points Given the reactions () and (2) below, determine A, HO for both HCI(g) and H0(g) all at 298 K. A,H 52.96 kJ/mol 4,He =-483.64 kJ/mol (a) , n® and , 1,0 for reaction (3), and (b) 2) 2H(g)+0(g)211:0g) 3) 4HI(g)+Oig)-212(s) + 2H 0(g) 2. 25 points Three moles o f an ideal gas (Cr-3 R) is expanded isothermally at 300 K from an initial (a) reversibly and (b) irreversibly and . , . AS, pressure of 10 Pa to a final pressure of 10 Pa in two ways: against a constant pressure equal to the final pressure of a gas of 105 Pa. Calculate q, w ASSur and ASot for each path. Assume that the surroundings are maintained at a constant tem AU, 300 K and pressure of 10 Pa throughout the process for either path. 3. 25 points For each of the following cases, calculate qg, w, AU, AS, ASsur and ASots and determi the process is spontaneous. In each case, assume that the process involves 3 moles of an ideal gas 3 R) that is initially at 600 K and 2-10% Pa. (a) The gas is cooled at constant volume corresponding to the initial volume to final temperature of 300 K. Assume the surroundings are maintained at 300 K temperature throughout the process. (b) The gas is cooled at constant pressure to final temperature of 300 K. Assume that the surroundings are maintained at temperature of 300 K throughout the process. (c) The gas undergoes a reversible, adiabatic expansion until volume is twice its initial value. 4. 20 points (a) Express for van der Waals gas and ideal gas in terms of the number of moles, gas constant, av volume of the gas, and van der Waals constant b (b) For an isothermal expansion, for which kind of gas (van der Waals gas versus ideal gas) will AS be greatest? 5. 10 points (a) State a criterion for spontaneity for a system that is allowed to interact with the surroundings in terms of the change in the entropy of the system and entropy of the surroundings (provide a mathematical statement) (b) The formulation of the Third Law of thermodynamics by M. Plank is following: "As T-0 K, S-0 for every chemically homogeneous substance in a perfect crystalline state." Explain why this formulation specifies the chemical and physical state of a substance and not only the temperatureExplanation / Answer
1. (a) dH of reaction (3) = 2 x - dH of reaction (1) + dH of reaction (2)
= 2 x (- 52.96) + (- 483.64)
= -589.56 kJ/mol.
dU of reaction (3) = dH of reaction (3) + dn x R x T
= -589.56 + (-3) x 8.314 x 298
= - 8022.276 kJ/mol.
(b) dHf of water (g) = dH of reaction (2) / 2
= - 241.82 kJ/mol.
2. (a) For reversible isothermal expansion,
w = - nRT ln(P1/P2) where, P1 and P2 are initial and final pressures.
= - 3 x 8.314 x 300 x ln (106/105)
= - 17229.32 kJ
q = -w = 17229.32 kJ.
dS = q / T = 17229.32 / 300 = 57.43 kJ/K.
dU = dH = 0 for an isothermal expansion as both depends on T and temperature is constant.
The simialr calculation is for the irreversible process also.
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