Holooly Plus Logo
Holooly Plus Logo

Question 12.10.2: Apply the Gibbs equilibrium phase rule to the reduction of c...

Apply the Gibbs equilibrium phase rule to the reduction of cupric oxide by ammonia and determine the number of degrees of freedom that exist for the reaction.

The Blue Check Mark means that this solution has been answered and checked by an expert. This guarantees that the final answer is accurate.
Learn more on how we answer questions.
3CuO_{(s)}+2NH_{3_{(g)}}=3Cu_{(s)}+3H_2O_{(g)}+N_{2(g)}

We use the equation

Φ + F = (C – r) + 2

Where:
Φ  is the number of phases in equilibrium
F  is the thermodynamic degrees of freedom
C  is the number of components
r  is the number of chemical reactions needed to specify equilibrium
2  accounts for the intensive variables pressure and temperature

For this case:
r   = 1
Φ   = 3 (2 solids and one gaseous)
3 + F   = (5 – 1) + 2 or F = 3

Related Answered Questions

During the chlorination of NiO in a reactor at 900 K, it is required that 90% conversion of the chlorine gas be achieved during a single pass through the reactor. Calculate the required total gas pressure.

Verified Answer:
The reaction is NiO_{(s)}+Cl_{2{(g)}}=NiCl_...

Consider the reduction of solid ZnO by CO to form Zn vapor and CO2 according to ZnO(s) + CO(g) = Zn(v) + CO2(g) For ZnO(s) = Zn(v) + ½O2(g): ΔG°(ii) = 460,200 −198T J For CO(g) + ½ O2(g) = CO2(g) : ΔG°(iii) = 282,400 + 86.81T J Summing ΔG°(ii) and ΔG°(iii) gives ΔG°(i) −177,800 −111.2T J The

Verified Answer:
a. At 1223 K, \Delta G^{\circ }_{(i)}=177,...

Compare the relative efficiencies of H2 and CO as reducing agents for metal oxides. For CO + ½O2 = CO2 , ΔG°(i) =−282,400+86.81T J and for H2 + ½O2 = H2O, ΔG°(ii) = −247,500 + 55.85T J

Verified Answer:
The values of ΔH° and ΔS° for these two reactions ...

Apply the Gibbs equilibrium phase rule to the oxidation of Cu to form cuprous oxide (Figure 12.1).

Verified Answer:
4Cu_{(s)}+O_{2_{(g)}}=2Cu_2O_{(s)} ...