Question

here is the full question nothing else was given

(2)(a)Solid copper(II) bromide, CuBr2(s), vaporizes via the following reaction:

2CuBr2(s)⇌2CuBr(s) + Br2(g)

Studies of the vapor pressure P of Br2(g) in equilibrium with CuBr2(s) and CuBr(s) show that it followsthe expression:P = (1.35 × 10^9bar) e^(–(11,642 K)/T) 320 K ≤ T ≤ 380 K. Find the equilibrium constant K, thereaction free energy change ΔG_r^o, the reaction enthalpy changeΔH_r^o,and the reaction entropy changeΔSrofor this reaction at T = 350 K.Hint:The K in the exponential is Kelvin where as the equilibrium constant is written as K.

(b) We would like to demonstrate that the appropriate free energy is minimized in thermodynamic equilibrium by considering a system at constant temperature and pressure, for which the Gibbs free energyG = U–TS + PV is appropriate

i) Write down the first law of thermodynamics linking heat, q, to ΔU and work.

ii)Write down an inequality relating thetemperature-weightedentropy change, TΔS to q.

iii)Use the result of (i) and (ii) to write down an inequality relating TΔS toΔUand work.

iv)Use the result of (iii) to obtain an inequality forΔG when the system is released out of equilibrium andundergoes spontaneous random changes.

v)Explain concisely why your result in (v) implies that the system has reached a state of thermal equilibrium if G has reached a minimum.

Answer 1

The First Law of Thermodynamics states that energy can be converted from one form to another with the interaction of heat, work and internal energy, but it cannot be created nor destroyed, under any circumstances. Mathematically, this is represented as

ΔU=q+w(1.1)(1.1)ΔU=q+w

with

ΔUΔU is the total change in internal energy of a system,
qq is the heat exchanged between a system and its surroundings, and
ww is the work done by or on the system.
Work is also equal to the negative external pressure on the system multiplied by the change in volume:

w=−pΔV(1.2)(1.2)w=−pΔV

where PP is the external pressure on the system, and ΔVΔV is the change in volume. This is specifically called "pressure-volume" work.

The internal energy of a system would decrease if the system gives off heat or does work. Therefore, internal energy of a system increases when the heat increases (this would be done by adding heat into a system). The internal energy would also increase if work were done onto a system. Any work or heat that goes into or out of a system changes the internal energy. However, since energy is never created nor destroyed (thus, the first law of thermodynamics), the change in internal energy always equals zero. If energy is lost by the system, then it is absorbed by the surroundings. If energy is absorbed into a system, then that energy was released by the surroundings:

ΔUsystem=−ΔUsurroundings(1.3)(1.3)ΔUsystem=−ΔUsurroundings

where ΔUsystem is the total internal energy in a system, and ΔUsurroundingsis the total energy of the surroundings.