HW Chp19 - Chapter 19: 8, 12, 14, 16; 20, 22, 24; 30, 32,...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
: 8, 12, 14, 16; 20, 22, 24; 30, 32, 33, 38, 40; 44, 48; 52, 54a, 54d, 56, 59, 62, 64-66; 72, 74, 76, 78, 80; 82, 84, 91a 19.1 (a) (b) Δ S is positive, because the disorder of the system increases. Each gas has greater motional freedom as it expands into the second bulb, and there are many more possible arrangements for the mixed gases. By definition, ideal gases experience no attractive or repulsive intermolecular interactions, so Δ H for the mixing of ideal gases is zero, assuming heat exchange only between the two bulbs. (c) The process is irreversible. It is inconceivable that the gases would reseparate. (d) The entropy change of the surroundings is related to Δ H for the system. Since we are mixing ideal gases and Δ H = 0, Δ H surr is also zero, assuming heat exchange only between the two bulbs. 19.2 (a) The process depicted is a change of state from a solid to a gas. Δ S increases because of the greater motional freedom of the particles. Δ H increases because both melting and boiling are endothermic processes. Since Δ G = Δ H – T Δ S, and both Δ H and Δ S are positive, the sign of Δ G depends on temperature. This is true for all phase changes. If the temperature of the system is greater than the boiling point of the substance, the process is spontaneous and Δ G is negative. If the temperature is lower than the boiling point, the process is not spontaneous and Δ G is positive. (b) If the process is spontaneous, the second law states that Δ S univ 0. Since Δ S sys increases, Δ S surr must decrease. If the change occurs via a reversible pathway, Δ S univ = 0 and Δ S surr = Δ S sys . If the pathway is irreversible, the magnitude of Δ S sys is greater than the magnitude of Δ S surr , but the sign of Δ S surr is still negative. 19.3 In the depicted reaction, both reactants and products are in the gas phase (they are far apart and randomly placed). There are twice as many molecules (or moles) of gas in the products, so Δ S is positive for this reaction. 19.4 (a) At 300 K, Δ H = T Δ S. Since Δ G = Δ H – T Δ S, Δ G = 0 at this point. When Δ G = 0, the system is at equilibrium. (b) The reaction is spontaneous when Δ G is negative. This condition is met when T Δ S > Δ H. From the diagram, T Δ S > Δ H when T > 300 K. The reaction is spontaneous at temperatures above 300 K. 19.5 (a) Analyze . The boxes depict three different mixtures of reactants and products for the reaction A 2 + B 2 2AB. Plan . Box 1 is an equilibrium mixture. By definition, Δ G = 0 for box 1. Calculate K and Δ G ° for the reaction from box 1. Boxes 2 and 3 are nonequilibrium mixtures. Calculate Q and Δ G for boxes 2 and 3. . ] B
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 8

HW Chp19 - Chapter 19: 8, 12, 14, 16; 20, 22, 24; 30, 32,...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online