Explain the difference between physical-equilibrium and chemical-equilibrium. Give two examples of each.

For which of the following reactions is $K_{\mathrm{c}}$ equal to $K_p$ ? (a) $4 \mathrm{NH}_3(g)+5 \mathrm{O}_2(g) \rightleftharpoons 4 \mathrm{NO}(g)+6 \mathrm{H}_2 \mathrm{O}(g)$

The equilibrium constant $K_{\mathrm{c}}$ for the reaction

$$\mathrm{I}_2(g) \rightleftharpoons 2 \mathrm{I}(g)$$

is $3.8 \times 10^{-5}$ at $727^{\circ} \mathrm{C}$. Calculate $K_{\mathrm{c}}$ and $K_P$ for the equilibrium

$$2 \mathrm{I}(g) \rightleftharpoons \mathrm{I}_2(g)$$

at the same temperature.

A reaction vessel contains $\mathrm{NH}_3, \mathrm{~N}_2$, and $\mathrm{H}_2$ at equilibrium at a certain temperature. The equilibrium concentrations are $\left[\mathrm{NH}_3\right]=0.25~M,\left[\mathrm{~N}_2\right]=0.11~M$, and $\left[\mathrm{H}_2\right]=1.91~M$. Calculate the equilibrium constant $K_{\mathrm{c}}$ for the synthesis of ammonia if the reaction is represented as (b) $\frac{1}{2} \mathrm{~N}_2(g)+\frac{3}{2} \mathrm{H}_2(g) \rightleftharpoons \mathrm{NH}_3(g)$

A reaction vessel contains $\mathrm{NH}_3, \mathrm{~N}_2$, and $\mathrm{H}_2$ at equilibrium at a certain temperature. The equilibrium concentrations are $\left[\mathrm{NH}_3\right]=0.25~M,\left[\mathrm{~N}_2\right]=0.11~M$, and $\left[\mathrm{H}_2\right]=1.91~M$. Calculate the equilibrium constant $K_{\mathrm{c}}$ for the synthesis of ammonia if the reaction is represented as (a) $\mathrm{N}_2(g)+3 \mathrm{H}_2(g) \rightleftharpoons 2 \mathrm{NH}_3(g)$

Consider this reaction:

$$\mathrm{N}_2(g)+\mathrm{O}_2(g) \rightleftharpoons 2 \mathrm{NO}(g)$$

If the equilibrium partial pressures of $\mathrm{N}_2, \mathrm{O}_2$, and $\mathrm{NO}$ are $0.15 \mathrm{~atm}, 0.33 \mathrm{~atm}$, and $0.050 \mathrm{~atm}$, respectively, at $2200^{\circ} \mathrm{C}$, what is $K_P$ ?

The equilibrium constant $K_P$ for the reaction

$$2 \mathrm{SO}_3(g) \rightleftharpoons 2 \mathrm{SO}_2(g)+\mathrm{O}_2(g)$$

is $5.0 \times 10^{-4}$ at $302^{\circ} \mathrm{C}$. What is $K_{\mathrm{c}}$ for this reaction?

What is the $K_P$ at $1273^{\circ} \mathrm{C}$ for the reaction

$$2 \mathrm{CO}(g)+\mathrm{O}_2(g) \rightleftharpoons 2 \mathrm{CO}_2(g)$$

if $K_{\mathrm{c}}$ is $2.24 \times 10^{22}$ at the same temperature?

Consider the following equilibrium process at $700^{\circ} \mathrm{C}$ :

$$2 \mathrm{H}_2(g)+\mathrm{S}_2(g) \rightleftharpoons 2 \mathrm{H}_2 \mathrm{~S}(g)$$

Analysis shows that there are $2.50$ moles of $\mathrm{H}_2, 1.35 \times$ $10^{-5}$ mole of $\mathrm{S}_2$, and $8.70$ moles of $\mathrm{H}_2 \mathrm{S}$ present in a $12.0-\mathrm{L}$ flask at equilibrium. Calculate the equilibrium constant $K_{\mathrm{c}}$ for the reaction.

The equilibrium constant $\left(K_{\mathrm{c}}\right)$ for the reaction

$$2 \mathrm{HCl}(g) \rightleftharpoons \mathrm{H}_2(g)+\mathrm{Cl}_2(g)$$

is $4.17 \times 10^{-34}$ at $25^{\circ} \mathrm{C}$. What is the equilibrium constant for the reaction

$$\mathrm{H}_2(g)+\mathrm{Cl}_2(g) \rightleftharpoons 2 \mathrm{HCl}(g)$$

at the same temperature?

Consider the following equilibrium system:

$$\mathrm{N}_2 \mathrm{O}_4(g) \rightleftharpoons 2 \mathrm{NO}_2(g) \quad \Delta H^{\circ}=58.0 \mathrm{~kJ} / \mathrm{mol}$$

(b) If the temperatures of the reacting system is changed at constant pressure, describe what a plot of $V$ versus $T$ would look like for the system.

Consider the following equilibrium system:

$$\mathrm{N}_2 \mathrm{O}_4(g) \rightleftharpoons 2 \mathrm{NO}_2(g) \quad \Delta H^{\circ}=58.0 \mathrm{~kJ} / \mathrm{mol}$$

(a) If the volume of the reacting system is changed at constant temperature, describe what a plot of $P$ versus $1 / V$ would look like for the system.

Iodine is sparingly soluble in water but much more so in carbon tetrachloride $\left(\mathrm{CCl}_4\right)$. The equilibrium constant, also called the partition coefficient, for the distribution of $\mathrm{I}_2$ between these two phases

$$\mathrm{I}_2(a q) \rightleftharpoons \mathrm{I}_2\left(\mathrm{CCl}_4\right)$$

is 83 at $20^{\circ} \mathrm{C}$. (c) Compare the result in (b) with a single extraction using $0.060 \mathrm{~L}$ of $\mathrm{CCl}_4$. Comment on the difference.

Iodine is sparingly soluble in water but much more so in carbon tetrachloride $\left(\mathrm{CCl}_4\right)$. The equilibrium constant, also called the partition coefficient, for the distribution of $\mathrm{I}_2$ between these two phases

$$\mathrm{I}_2(a q) \rightleftharpoons \mathrm{I}_2\left(\mathrm{CCl}_4\right)$$

is 83 at $20^{\circ} \mathrm{C}$. (b) The student now repeats the extraction of $\mathrm{I}_2$ with another $0.030 \mathrm{~L}$ of $\mathrm{CCl}_4$. Calculate the fraction of the $\mathrm{I}_2$ from the original solution that remains in the aqueous phase.