Extra Credit 30

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Q13.4.16

Calculate the number of moles of HI that are at equilibrium with 1.25 mol of H₂ and 1.25 mol of I₂ in a 5.00−L flask at 448 °C.

\(\ce{H2 + I2 ⇌ 2HI} \hspace{20px} K_c=\textrm{50.2 at 448 °C}\)

Solution

We first calculate the concentration of H_2 and I_2

C_H2=C_I2=1.25 mol/ 5.00 L= 0.25 mol/L

Then, we set the concentration of HI at equilibrium is x.

Use the Ice table:

K_C=50.2=x^2/[(0.25-x)(0.25-x)] x = 0.219

So, the number of moles of HI is 0.219mol/L * 5.00 L= 1.095 mol

Q14.6.5

What is the pH of 1.000 L of a solution of 100.0 g of glutamic acid (C₅H₉NO₄, a diprotic acid; K₁ = 8.5 × 10⁻⁵, K₂ = 3.39 × 10⁻¹⁰) to which has been added 20.0 g of NaOH during the preparation of monosodium glutamate, the flavoring agent? What is the pH when exactly 1 mol of NaOH per mole of acid has been added?

Q16.4.34

An important source of copper is from the copper ore, chalcocite, a form of copper(I) sulfide. When heated, the Cu₂S decomposes to form copper and sulfur described by the following equation:

\[\ce{Cu2S}(s)⟶\ce{Cu}(s)+\ce{S}(s)\]

(a) Determine \(ΔG^\circ_{298}\) for the decomposition of Cu₂S(s).
The reaction of sulfur with oxygen yields sulfur dioxide as the only product. Write an equation that describes this reaction, and determine \(ΔG^\circ_{298}\) for the process.
The production of copper from chalcocite is performed by roasting the Cu₂S in air to produce the Cu. By combining the equations from Parts (a) and (b), write the equation that describes the roasting of the chalcocite, and explain why coupling these reactions together makes for a more efficient process for the production of the copper.

Solution

a. Find the \(ΔG^\circ_{298}\) for Cu2S in the Standard Thermodynamic Table is -86.2 KJ/mol, and we know the \(ΔG^\circ_{298}\) for the pure substances is 0.

So, the \(ΔG^\circ_{298}\) for the decomposition of Cu2S is 0+0-(-86.2)=86.2 KJ/mol.

b. The equation of the formation of sulfur dioxide is \[\ce{S}(s)+\ce{O2}(g)⟶\ce{SO2}(g)\]Again, Find the \(ΔG^\circ_{298}\) of SO2, which is -300.1 KJ/mol.

So, the \(ΔG^\circ_{298}\) for the formation of sulfur dioxide is -300.1-(0+0)=-300.1 KJ/mol.

c. The equation of roasting the Cu2S in air is \[\ce{Cu2S}(s)+\ce{O2}(g)⟶\ce{Cu}(s)+\ce{SO2}(g)\]Combining these reactions together makes a more efficient process is because the reaction \[\ce{Cu2S}(s)⟶\ce{Cu}(s)+\ce{S}(s)\] is a reversible reaction. Therefore, if we wanted to obtain more copper, we should remove the another product sulfur, and the best way to remove sulfur is to form sulfur dioxide. Consequently, we can get a more efficient process for the production of copper.

Q15.3.7

Calculate the Co²⁺ equilibrium concentration when 0.100 mole of [Co(NH₃)₆](NO₃)₂ is added to a solution with 0.025 M NH₃. Assume the volume is 1.00 L.

Solution

Firstly, we can find the equilibrium constant of formation of [Co(NH₃)₆]²⁺ , which is 5.0*10^4.

Then, we set the Δ concentration of NH3 at equilibrium is x M.

Use the ICE table, we can get the following equation:

K_c=(0.1-1/6x)/[(1/6x)(0.025+x)^6]=50000 x=0.1679

So, the equilibrium concentration of Co²⁺ is 1/6x=0.1679/6=0.028 M

Q16.4.31

In glycolysis, the reaction of glucose (Glu) to form glucose-6-phosphate (G6P) requires ATP to be present as described by the following equation:

\[\mathrm{Glu + ATP ⟶ G6P + ADP} \hspace{20px} ΔG^\circ_{298}=\mathrm{−17\: kJ}\]

In this process, ATP becomes ADP summarized by the following equation:

\[\mathrm{ATP⟶ADP} \hspace{20px} ΔG^\circ_{298}=\mathrm{−30\: kJ}\]

Determine the standard free energy change for the following reaction, and explain why ATP is necessary to drive this process:

\[\mathrm{Glu⟶G6P} \hspace{20px} ΔG^\circ_{298}=\:?\]

Solution:

According to the Hess's Law

\[\mathrm{Glu + ATP ⟶ G6P + ADP}\+\mathrm{ADP⟶ATP}\=\mathrm{Glu⟶G6P}\]

So, the \(ΔG^\circ_{298}\) for Glu⟶G6P equils to -17+-(-30)=13KJ

Glucose is a relatively stable substance. In this problem, we want to add a big functional group: ·H₂PO₃ (This is the P), so we need such functional group be a active state. When the weakest P on the ATP molecular releases from the ATP, it is on active state. Therefore, it will react with a frank molecular, Glucose. In order to get such a active P, the ATP is necessary.