Chapter 8.15, Problem 64SEP

Using Figure P8.40, explain what the phase diagram is showing when the overall alloy composition is A1–43wt% Ni (below 854°C). Why is there a vertical line at that point in the phase diagram? Verify that the formula for the compound is Al₃Ni. What do you call such a compound?

To determine

Using Figure P8.40, explain what the phase diagram is showing when the overall alloy composition is Al–43wt% Ni (below 854°C). Why is there a vertical line at that point in the phase diagram. Verify that the formula for the compound is Al3Ni. What do you call such a compound.

Explanation of Solution

Express the number of moles in nickel.

 $N_{\text{Ni}}=w{t}_{\text{Ni}}\left[\frac{\text{1}\text{mol}}{{\text{M}}_{\text{Ni}}}\right]$                                                                           (I)

Here, number of moles in nickel is $N_{\text{Ni}}$, weight percentage of nickel in compound is $w_{\text{Ni}}$ and atomic mass of nickel is ${\text{M}}_{\text{Ni}}$.

Express the number of moles in aluminum.

 $N_{\text{Al}}=w{t}_{\text{Al}}\left[\frac{\text{1}\text{mol}}{{\text{M}}_{\text{Al}}}\right]$                                                                           (II)

Here, number of moles in aluminum is $N_{\text{Al}}$, weight percentage of aluminum in compound is $w_{\text{Al}}$ and atomic mass of aluminum is ${\text{M}}_{\text{Al}}$.

Express mole percentage of aluminum.

 $\text{mol\%}\text{Al}=\frac{N_{\text{Al}}}{N_{\text{Al}}+N_{\text{Ni}}}$                                                                      (III)

Express mole percentage of nickel.

 $\text{mol\%}\text{Ni}=\frac{N_{\text{Ni}}}{N_{\text{Al}}+N_{\text{Ni}}}$                                                                      (IV)

Conclusion:

The phase diagram shown in Figure P8.40 is presenting the formation or development of an intermetallic compound between aluminum and nickel as Al–43wt% Ni indicates some substitution between them.

The vertical line at that composition signifies a stoichiometric behavior of the formed intermetallic compound.

Verification of formula:

To verfify the formula for the compound, let the mass of compound be $\text{100}\text{g}$. Thus, the weight percentage of of nickel is 43wt% and that of alumium is 57wt%.

 $\begin{array}{l}{w}_{\text{Ni}}=43\text{wt\%}\\ w_{\text{Al}}=57\text{wt\%}\end{array}$

Write the atomic mass of nickel and aluminum as below:

 $\begin{array}{l}{\text{M}}_{\text{Ni}}=58.68\text{g}/\text{mol}\\ {\text{M}}_{\text{Al}}=26.98\text{g}/\text{mol}\end{array}$

Substitute $\text{43}\text{g}$ for $w_{\text{Ni}}$ and $58.68\text{g}/\text{mol}$ for ${\text{M}}_{\text{Ni}}$ in Equation (I).

 $\begin{array}{c}{N}_{\text{Ni}}=\left(43\text{g}\right)\left[\frac{\text{1}\text{mol}}{58.68\text{g}/\text{mol}}\right]\\ =0.73\end{array}$

Substitute $\text{57}\text{g}$ for $w_{\text{Al}}$ and $26.98\text{g}/\text{mol}$ for ${\text{M}}_{\text{Al}}$ in Equation (II).

 $\begin{array}{c}{N}_{\text{Al}}=\left(\text{57}\text{g}\right)\left[\frac{\text{1}\text{mol}}{26.98\text{g}/\text{mol}}\right]\\ =2.11\end{array}$

Substitute $0.73$ for $N_{\text{Ni}}$ and $2.11$ for $N_{\text{Al}}$ in Equation (III).

 $\begin{array}{c}\text{mol\%}\text{Al}=\frac{2.11}{2.11+0.73}\\ =0.75\\ =75\%\end{array}$

Substitute $0.73$ for $N_{\text{Ni}}$ and $2.11$ for $N_{\text{Al}}$ in Equation (IV).

 $\begin{array}{c}\text{mol\%}\text{Ni}=\frac{0.73}{2.11+0.73}\\ =0.25\\ =25\%\end{array}$

For each nickel atom, there are 3 atoms of aluminum, hence the formula is ${\text{Al}}_{\text{3}}\text{Ni}$ and such kind of compounds are called as intermetallics as there is a compound between two metals.