Reaction 1:

CuO(s)→Cu(s)+12O2(g)ΔG°=155kJ/molrxn


A chemist wants to produce Cu(s) from a sample of pure CuO(s) according to reaction 1, represented by the equation above.


(a) Using the data in the following table, calculate the value of the standard entropy change, ΔS°, for the reaction.


Substance Absolute Entropy at298 K (JK−1mol−1)

Cu(s) 33

O2(g) 205

CuO(s) 43


(b) Given that ΔG° for reaction 1 is positive (155kJ/molrxn), what must be true about the sign of ΔH° for the reaction? Justify your answer.


The decomposition of CuO(s) into Cu(s) and O2(g) is not a thermodynamically favored reaction. However, to produce Cu(s) from CuO(s), the chemist decides to pass H2(g) gas over the CuO(s) as it is heated strongly, as represented below.


Reaction 2:

CuO(s)+H2(g)−----heat→ Cu(s)+H2O(g)


(c) Reaction 2 is what type of chemical reaction? Justify your answer.


A third chemical reaction is represented below.


Reaction 3:

H2(g)+12O2(g)→H2O(g)ΔG°=−229kJ/molrxn


(d) Show how a combination of reaction 1 and reaction 3 can be used to produce reaction 2.


(e) Determine the value of ΔG° for reaction 2.


The chemist takes the Cu(s) produced from reaction 2 and uses it to make an electrode in a galvanic cell, as shown in the following diagram.


The figure presents a diagram of a galvanic cell consisting of two solutions in beakers, two metal electrodes, a salt bridge, and a wire. The half-cell on the left shows a copper electrode partially submerged in a solution of 1.0 molar copper sulfate. The half-cell on the right shows a silver electrode partially submerged in a solution of 1.0 molar silver nitrate. A salt bridge connects the two solutions. A wire connects the copper electrode and the silver electrode, and an arrow indicates that the electron flow through the wire is from the copper to the silver.


(f) Using the reduction half-reactions in the table below, write a balanced net-ionic equation representing the overall reaction that takes place as the cell operates.


Half-Reaction E° (volts)

Cu2+(aq)+2e−→Cu(s) 0.337

Ag+(aq)+e−→Ag(s) 0.800


(g) Determine the value of E° for the cell.

(h) Determine the value of ΔG° for the cell reaction.

Reaction 1CuOsCus12O2gΔG155kJmolrxnA chemist wants to produce Cus from a sample of pure CuOs according to reaction 1 represented by the equation abovea Using th class=
Reaction 1CuOsCus12O2gΔG155kJmolrxnA chemist wants to produce Cus from a sample of pure CuOs according to reaction 1 represented by the equation abovea Using th class=
Reaction 1CuOsCus12O2gΔG155kJmolrxnA chemist wants to produce Cus from a sample of pure CuOs according to reaction 1 represented by the equation abovea Using th class=
Reaction 1CuOsCus12O2gΔG155kJmolrxnA chemist wants to produce Cus from a sample of pure CuOs according to reaction 1 represented by the equation abovea Using th class=
Reaction 1CuOsCus12O2gΔG155kJmolrxnA chemist wants to produce Cus from a sample of pure CuOs according to reaction 1 represented by the equation abovea Using th class=

Respuesta :

a. ΔS°, for the reaction

= 3252-1032

= 2220 K (JK−1mol−1).

b. ΔG= ΔH- TΔS

155x10^3 = ΔH- 222ox 298

= ΔH= 816.56 KJ.

c. CuO(s)+H2(g)−----heat→ Cu(s)+H2O(g)

balance eqn -  2CuO(s)+2H2(g)−----heat→ 2Cu(s)+2H2O(g)

It is displacement reaction.

H2(g)+12O2(g)→H2O(g)ΔG°=−229kJ/molrxn

d. Adding 1 and 3 eqn to get 2 eqn.

CuO(s)→Cu(s)+12O2(g)----Cu(s)+12O2(g)+ H2O(g)

e. ΔG = ΔG reaction 1 - ΔG reaction 3

= 155-229

= - 74 KJ/mol.

f. Overall cell reaction :

2Ag +Cu ⇒ 2Ag +Cu(s)2+

There will be greater reduction potential so Ag will have reduction potential while Cu will have oxidation.

g. E= -0.337 + 0.8

= 0.463 V

h.  ΔG° = -89.359 kJ.

What is entropy ?

It is the amount of energy unavailable to do work in a system.

Thus it is well explained.

To learn more about the Entropy  refer to link ;

https://brainly.com/question/6364271

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