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2 answers:
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Answer:
ΔG° = -80.9 KJ
Assuming this reaction takes place at room temperature (25 °C):
K=
Explanation:
1) Reduction potentials
First of all one should look up the reduction potentials for the species envolved:
→
E°red=1.61V
→
E°red=0.771V
2) Redox pair
Knowing their reduction pontentials one can determine a redox pair: one species must oxidate while the other is reducing. <u>Remember: the table gives us the reduction potential, so if we want to know the oxidation potential all that has to be done is reverce the equation and change the potencial signal (multiply to -1).</u>
1) reduces while
oxidates
(oxidation) →
E°oxi=-0.771V
(reduction) →
E°red=1.61V
(overall equation) →
E°=Ereduction + Eoxidation= 1.61 v+(-0.771 v) = 0.839v
The cell potential can also be calculated as the cathode potencial minus the anode potential:
E° = E cathode - E anode =1.61 v - 0.771 v=0.839 v
3) Gibbs free energy and Equilibrium constant
ΔG°=-nFE°, where 'n' is the number of electrons involved in the redox equation, in this case n is 1. 'F' is the Faraday constant, whtch is 96500 C. E° is the standard cell potencial.
ΔG°=-nFE°=-1*96500*0.839
ΔG° = - 80963 J = -80.9 KJ
The Nerst equation gives us the relation of chemical equilibrium and Electric potential.
E=E°-
Where 'R' is the molar gas constant (8.314 J/mol)
It's known that in the equilibrium E=0, so the Nerst equation, at equilibrium, becomes:
E°=
Isolating for 'K' gives:
This shows that 'K' is a fuction of temperature. Assuming this reaction takes place at room temperature (25 °C):
K=
Answer:
1) Endothermic.
2)
3)
Explanation:
Hello there!
1) In this case, for these calorimetry problems, we can realize that since the temperature decreases the reaction is endothermic because it is absorbing heat from the solution, that is why the temperature goes from 22.00 °C to 16.0°C.
2) Now, for the total heat released by the reaction, we first need to assume that all of it is released by the solution since it is possible to assume that the calorimeter is perfectly isolated. In such a way, it is also valid to assume that the specific heat of the solution is 4.184 J/(g°C) as it is mostly water, therefore, the heat released by the reaction is:
3) Finally, since the enthalpy of reaction is calculated by dividing the heat released by the reaction over the moles of the solute, in this case NH4Cl, we proceed as follows:
Best regards!
Best regards!