According to advertisements, “a diamond is forever.”

(a) Calculate $\Delta H^{\circ}, \Delta S^{\circ},$ and $\Delta G^{\circ}$ at 298 $\mathrm{K}$ for the phase change

(b) Given the conditions under which diamond jewelry is normally kept, argue for and against the statement in the ad.

(c) Given the answers in part (a), what would need to be done to make synthetic diamonds from graphite?

(d) Assuming $\Delta H^{\circ}$ and $\Delta S^{\circ}$ do not change with temperature, can graphite be converted to diamond spontaneously at 1 $\mathrm{atm} ?$

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A solution is prepared by adding 0.10 mole of $\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6} \mathrm{Cl}_{2}$ to 0.50 $\mathrm{L}$ of $3.0 M$ $\mathrm{NH}_{3}$ . Calculate $\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\right]$ and $\left[\mathrm{Ni}^{2+}\right]$ in this solution. $K_{\text { overall }}$ for $\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}$ is $5.5 \times 10^{8} .$ That is,

$$5.5 \times 10^{8}=\frac{\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\right]}{\left[\mathrm{Ni}^{2+}\right]\left[\mathrm{NH}_{3}\right]^{6}}$$

for the overall reaction

$$\mathrm{Ni}^{2+}(a q)+6 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}(a q)$$

Calculate the equilibrium concentration of $\mathrm{Ni}^{2+}$ in a $1.0-M$ solution $\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]\left(\mathrm{NO}_{3}\right)_{2}$

Consider the reaction:

$$\mathrm{NH}_{4} \mathrm{HS}(s) \rightleftharpoons \mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{S}(g)$$

An equilibrium mixture of this reaction at a certain temperature has $\left[\mathrm{NH}_{3}\right]=0.278 \mathrm{M}$ and $\left[\mathrm{H}_{2} \mathrm{S}\right]=0.355 \mathrm{M} .$ What is the value of

the equilibrium constant $\left(K_{\mathrm{c}}\right)$ at this temperature?

In the presence of $\mathrm{NH}_{3}, \mathrm{Cu}^{2+}$ forms the complex ion $\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+} .$ If the equilibrium concentrations of $\mathrm{Cu}^{2+}$ and $\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}$ are $1.8 \times 10^{-17} \mathrm{M}$ and $1.0 \times 10^{-3} \mathrm{M},$ respectively, in a $1.5-M \mathrm{NH}_{3}$ solution, calculate the value for the overall formation constant of $\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}.$

$$\mathrm{Cu}^{2+}(a q)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q) \qquad K_{\mathrm{overall}}=?$$

The $K_{b}$ for methylamine $\left(\mathrm{CH}_{3} \mathrm{NH}_{2}\right)$ at $25^{\circ} \mathrm{C}$ is given in Appendix $\mathrm{D}$ . (a) Write the chemical equation for the equilibrium that corresponds to $K_{b}$ . (b) By using the value of $K_{b},$ calculate $\Delta G^{\circ}$ for the equilibrium in part (a). (c) What is the value of $\Delta G$ at equilibrium? (d) What is the value of $\Delta G$ when $\left[\mathrm{H}^{+}\right]=6.7 \times 10^{-9} M,\left[\mathrm{CH}_{3} \mathrm{NH}_{3}^{+}\right]=2.4 \times 10^{-3} \mathrm{M}$ and $\left[\mathrm{CH}_{3} \mathrm{NH}_{2}\right]=0.098 \mathrm{M} ?$