Thermodynamics — Practice

Question 1

For N2 + 3H2 -> 2NH3, delta-H = -92.4 kJ at 298 K. Find delta-U given delta-n_g = -2.

Accepted Answer

delta-H = delta-U + delta-n_g RT. -92.4 = delta-U + (-2)(8.314 x 298)/1000 = delta-U - 4.96. So delta-U = -87.44 kJ.

Question 2

Calculate delta-fH of C2H6 given delta-cH(C) = -393.5, delta-fH(H2O) = -285.8, and delta-cH(C2H6) = -1560 kJ/mol.

Accepted Answer

delta-fH(C2H6) = 2(-393.5) + 3(-285.8) - (-1560) = -787 - 857.4 + 1560 = -84.4 kJ/mol.

Question 3

For a reaction delta-H = 30 kJ and delta-S = 100 J/K. Find the temperature above which it becomes spontaneous.

Accepted Answer

At equilibrium delta-G = 0, so T = delta-H/delta-S = 30000/100 = 300 K. Since delta-H and delta-S are both positive, the reaction is spontaneous above 300 K.

Question 4

Is bond breaking endothermic or exothermic, and what about bond formation?

Accepted Answer

Bond breaking is endothermic (energy must be supplied); bond formation is exothermic (energy is released).

Question 5

What is the significance of delta-G = 0?

Accepted Answer

delta-G = 0 means the system is at equilibrium -- there is no net change and the forward and reverse rates are equal.

Practice — Thermodynamics

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