Calculate the equilibrium composition of a Cu-Ni alloy (in wt%) capable of existing in equilibrium with CU2O and NiO at 727°C (1000 K) from the following data: At 727°C (lOOOK) - AG°Cu Q = -18,230 cal/mole (-76,274 J/mol), and AG°Ni0 = ~ 3 0 » 1 5 0 cal/mole (-126,148 J/mol). Atomic weights of Cu and Ni are 63.54 and 58.7 respectively.
Question
Calculate the equilibrium composition of a Cu-Ni alloy (in wt%) capable of existing in equilibrium with CU2O and NiO at 727°C (1000 K) from the following data: At 727°C (lOOOK) - AG°Cu Q = -18,230 cal/mole (-76,274 J/mol), and AG°Ni0 = ~ 3 0 » 1 5 0 cal/mole (-126,148 J/mol). Atomic weights of Cu and Ni are 63.54 and 58.7 respectively.
Solution
To calculate the equilibrium composition of a Cu-Ni alloy, we need to use the Gibbs free energy change (ΔG) for the formation of Cu2O and NiO. The Gibbs free energy change for a reaction is given by the equation:
ΔG = ΔH - TΔS
where ΔH is the enthalpy change, T is the temperature in Kelvin, and ΔS is the entropy change. However, in this case, we are given the values of ΔG for the formation of Cu2O and NiO directly.
The equilibrium constant (K) for a reaction is related to the Gibbs free energy change by the equation:
ΔG = -RT ln K
where R is the gas constant (8.314 J/mol K), T is the temperature in Kelvin, and K is the equilibrium constant. We can rearrange this equation to solve for K:
K = e^(-ΔG/RT)
We can calculate K for the formation of Cu2O and NiO using the given values of ΔG and the temperature (1000 K):
K_Cu2O = e^(-(-76,274 J/mol) / (8.314 J/mol K * 1000 K)) = 2.68 x 10^16
K_NiO = e^(-(-126,148 J/mol) / (8.314 J/mol K * 1000 K)) = 1.18 x 10^24
The equilibrium composition of the Cu-Ni alloy can then be calculated using the equilibrium constants and the initial composition of the alloy. However, the initial composition of the alloy is not given in the question, so we cannot calculate the equilibrium composition.
Similar Questions
Is it possible to have a copper–nickel alloy that, at equilibrium, con-sists of an a phase of composition 37 wt% Ni–63 wt% Cu and also a liquid phase of composition20 wt% Ni–80 wt% Cu? If so, what will be the approximate temperature of the alloy? If this isnot possible, explain why
Solid Au-Cu alloys areregular intheir thermodynamic behaviour. The integral enthalpy of mixing, ΔΗ , at 447°C (720 K) is given below as a function of composition. xCu : ΔΗΜ, cal/mole : 0.2 0.3 -585 -850 0.4 -1075 -4498 0.5 -1220 -5104) 4 (kïf , J/mol : -2448 -3556
A copper–nickel alloy of composition 70 wt% Ni–30 wt% Cu isslowly heated from a temperature of 1300C (2370F).(a) At what temperature does the first liquid phase form?(b) What is the composition of this liquid phase?(c) At what temperature does complete melting of the alloy occur?(d) What is the composition of the last solid remaining prior to complete melting?
2Cu+ (aq) → Cu (s) + Cu+ (aq) 2Cu+ (𝑎𝑞) → Cu (𝑠) + Cu+ (𝑎𝑞) is +0.36 V at 298 K. The equilibrium constant of the reaction is
Of the metals Zn, Mg and Fe, which removes copper (II) ion from the solution mostcompletely? The following equilibrium constants are measured at room temperature.Zn (s) + Cu2+ (aq) → Cu(s) + Zn2+(aq) Kc = 2 x 1037Mg (s) + Cu2+ (aq) → Cu(s) + Mg2+(aq) Kc = 2 x 1090Fe (s) + Cu2+ (aq) → Cu(s) + Zn2+(aq) Kc = 2 x 1026Give reason for your answer
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.