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Predict the sign of the entropy change for the following processes. Give a reason for your prediction.(a) $\mathrm{Pb}^{2+}(a q)+\mathrm{S}^{2-}(a q) \longrightarrow \mathrm{PbS}(s)$(b) $2 \mathrm{Fe}(s)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{Fe}_{2} \mathrm{O}_{2}(s)$(c) $2 \mathrm{C}_{6} \mathrm{H}_{14}(l)+19 \mathrm{O}_{2}(g) \longrightarrow 14 \mathrm{H}_{2} \mathrm{O}(g)+12 \mathrm{CO}_{2}(g)$

a. negative.b. negative.c. positive.

Chemistry 102

Chapter 16

Thermodynamics

University of Central Florida

University of Maryland - University College

Lectures

03:07

A liquid is a nearly incom…

04:38

A liquid is a state of mat…

02:24

Predict whether the entrop…

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Calculate the standard ent…

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Predict the sign of the en…

02:40

Arrange the entropy change…

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02:09

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Calculate the standard mol…

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Indicate whether entropy i…

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Use $S^{\circ}$ values to …

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Use the standard molar ent…

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02:46

we know that mentality is a state function on dhe. It is the energy change at constant pressure. During the course off reaction, we know that the energy for gaseous state the ant elope for gosh, a state will be the highest as compared to the lowest, which is solid state for option E. We have conversion from acquis to Saul it in this case and no ji is being released during the reaction. The precipitation decreases the number off free irons in the solution hence and and mentality will be negative. In option B, we have changed off mentality from gash estate for oxygen in tow. Solid final state. Over here, the energy is released during the reaction. So the sign off an talebi will be negative again. For option C, we have a liquid state on gas estate as the reactant giving us the product for gaseous state. Hence, there is an increase off seven moles off gas from the reactant liquid. Thus, the sign off mentality will be positive. In this case,

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