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11111nata11111 [884]
3 years ago
6

Heating copper(II) oxide at 400°C does not produce any appreciable amount of Cu: CuO(s) ⇆ Cu(s) + 1 2 O2(g) ΔG o = 127.2 kJ/mol

However, if this reaction is coupled to the conversion of graphite to carbon monoxide, it becomes spontaneous. Write an equation for the coupled process and calculate the equilibrium constant for the coupled reaction. Be sure to include the states of the chemical species.
Chemistry
2 answers:
Allisa [31]3 years ago
7 0

Explanation:

The reaction is as follows.

          CuS \rightleftharpoons Cu(s) + \frac{1}{2}O_{2}(g)

As, value of \Delta G is positive. Therefore, reaction is non-spontaneous.

     CuO(s) + C(graphite) \rightleftharpoons Cu(s) + CO(g)

     \Delta G = \sum \Delta G_{f}_{product} - \sum \Delta G_{f}_{reactant}

                  = [-137.2 - (127.2 kJ/mol)]

                  = -10 kJ/mol

Since, value of \Delta G is negative here so, reaction is spontaneous.

Also,  \Delta G = -RT ln

where,        R = 8.314 J/mol K

                  T = 400^{o}C = (400 +  273) K = 673 K

                  K = equilibrium constant

       -10 \times 10^{3} J/mol = -8.314 J/mol K \times 673 ln K

           100 = 2.303 \times 8.314 J/mol K \times 673 log K

           log K = 0.00776

              K = 1.018

Therefore, we can conclude that equilibrium constant for the coupled reaction is 1.018.

loris [4]3 years ago
6 0

Answer:

CuO(s)Cu(s)+\frac{1}{2}O_2(g);\Delta G^o_1=127.2kJ/mol \\2C(s)+O_2(g)2CO(s);\Delta G^o_2=-137.16kJ/mol

K=5.93

Explanation:

Hello,

In this case, the coupled process contains the following two chemical reactions:

CuO(s)Cu(s)+\frac{1}{2}O_2(g);\Delta G^o_1=127.2kJ/mol \\2C(s)+O_2(g)2CO(s);\Delta G^o_2=-137.16kJ/mol

By taking the total Gibbs free energy for this coupled reactions:

\Delta G^o_{T}=127.2kJ/mol-137.16kJ/mol=-9.96kJ/mol

In such a way, we compute the equilibrium constant as follows:

K=exp(-\frac{\Delta G^o_{T}}{RT} )=exp(-\frac{(-9960J/mol)}{8.314J/molK*673.15} )\\K=5.93

Best regards.

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Taking into account the reaction stoichiometry, 8 moles of Ag can be produced from 8 moles of AgNO₃ and 5 moles of Zn.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

2 AgNO₃ + Zn → 2 Ag + Zn(NO₃)₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

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To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 1 mole of Zn reacts with 2 moles of AgNO₃, 5 moles of Zn reacts with how many moles of AgNO₃?

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But 10 moles of AgNO₃ are not available, 8 moles are available. Since you have less moles than you need to react with 5 moles of Zn, AgNO₃ will be the limiting reagent.

<h3>Moles of Ag formed</h3>

Considering the limiting reagent, the following rule of three can be applied: if by reaction stoichiometry 2 moles of AgNO₃ form 2 moles of Ag, 8 moles of AgNO₃ form how many moles of Ag?

amount of moles of Ag=\frac{8 moles of AgNO_{3}x2 moles of Ag }{2 moles of AgNO_{3}}

<u><em>amount of moles of Ag= 8 moles</em></u>

Then, 8 moles of Ag can be produced from 8 moles of AgNO₃ and 5 moles of Zn.

Learn more about the reaction stoichiometry:

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