Equilibrium

Given: A liquid–vapour equilibrium in a sealed container at fixed temperature; volume is suddenly increased.

(a) Initial effect on vapour pressure:
When the volume is suddenly increased, the same number of vapour molecules now occupy a larger volume. Therefore, the concentration (and hence the partial pressure) of the vapour decreases initially.

(b) Initial change in rates:
- Rate of evaporation: Evaporation depends on the nature of the liquid and temperature, not on the volume of the container. Hence, the rate of evaporation remains unchanged initially.
- Rate of condensation: Condensation depends on the concentration (number density) of vapour molecules. Since the vapour pressure has decreased, the rate of condensation decreases initially.

Because rate of evaporation > rate of condensation, more liquid evaporates to restore equilibrium.

(c) Final state after equilibrium is restored:
More liquid evaporates until the rate of evaporation once again equals the rate of condensation. At the new equilibrium, the vapour pressure equals the original vapour pressure (since temperature is unchanged and vapour pressure depends only on temperature). Thus, the final vapour pressure is the same as the original vapour pressure.

Given: Total pressure $P = 10^5\,\mathrm{Pa}$; iodine vapour contains 40% by volume of I atoms.

Concept: % by volume = % by moles (for ideal gases).

So mole fraction of I atoms, $\chi_{\mathrm{I}} = 0.40$
Mole fraction of $\mathrm{I}_2$, $\chi_{\mathrm{I}_2} = 1 - 0.40 = 0.60$

Partial pressures:
$$p_{\mathrm{I}} = 0.40 \times 10^5 = 4.0 \times 10^4\,\mathrm{Pa}$$
$$p_{\mathrm{I}_2} = 0.60 \times 10^5 = 6.0 \times 10^4\,\mathrm{Pa}$$

$K_p$ expression:
$$K_p = \frac{(p_{\mathrm{I}})^2}{p_{\mathrm{I}_2}} = \frac{(4.0 \times 10^4)^2}{6.0 \times 10^4}$$
$$= \frac{16 \times 10^8}{6.0 \times 10^4} = \frac{16}{6} \times 10^4$$
$$\boxed{K_p = 2.67 \times 10^4\,\mathrm{Pa}}$$

Concept: Pure solids and pure liquids are excluded from the equilibrium constant expression. Only gaseous and aqueous species are included.

(i) $2\mathrm{NOCl}(\mathrm{g}) \rightleftharpoons 2\mathrm{NO}(\mathrm{g}) + \mathrm{Cl}_2(\mathrm{g})$
$$K_c = \frac{[\mathrm{NO}]^2[\mathrm{Cl}_2]}{[\mathrm{NOCl}]^2}$$

(ii) $2\mathrm{Cu(NO_3)_2(s)} \rightleftharpoons 2\mathrm{CuO(s)} + 4\mathrm{NO_2(g)} + \mathrm{O_2(g)}$

Solids are excluded:
$$K_c = [\mathrm{NO_2}]^4[\mathrm{O_2}]$$

(iii) $\mathrm{CH_3COOC_2H_5(aq)} + \mathrm{H_2O(l)} \rightleftharpoons \mathrm{CH_3COOH(aq)} + \mathrm{C_2H_5OH(aq)}$

Pure liquid water is excluded:
$$K_c = \frac{[\mathrm{CH_3COOH}][\mathrm{C_2H_5OH}]}{[\mathrm{CH_3COOC_2H_5}]}$$

(iv) $\mathrm{Fe^{3+}(aq)} + 3\mathrm{OH^-(aq)} \rightleftharpoons \mathrm{Fe(OH)_3(s)}$

Solid $\mathrm{Fe(OH)_3}$ is excluded:
$$K_c = \frac{1}{[\mathrm{Fe^{3+}}][\mathrm{OH^-}]^3}$$

(v) $\mathrm{I_2(s)} + 5\mathrm{F_2(g)} \rightleftharpoons 2\mathrm{IF_5(g)}$

Solid $\mathrm{I_2}$ is excluded:
$$K_c = \frac{[\mathrm{IF_5}]^2}{[\mathrm{F_2}]^5}$$

Formula: $K_p = K_c(RT)^{\Delta n}$, so $K_c = \dfrac{K_p}{(RT)^{\Delta n}}$

where $\Delta n$ = (moles of gaseous products) $-$ (moles of gaseous reactants), $R = 0.0831\,\mathrm{bar\,L\,mol^{-1}\,K^{-1}}$.

(i) $2\mathrm{NOCl}(\mathrm{g}) \rightleftharpoons 2\mathrm{NO}(\mathrm{g}) + \mathrm{Cl}_2(\mathrm{g})$

$\Delta n = (2+1) - 2 = 1$

$$K_c = \frac{K_p}{(RT)^1} = \frac{1.8 \times 10^{-2}}{0.0831 \times 500} = \frac{1.8 \times 10^{-2}}{41.55}$$
$$\boxed{K_c = 4.33 \times 10^{-4}}$$

(ii) $\mathrm{CaCO_3(s)} \rightleftharpoons \mathrm{CaO(s)} + \mathrm{CO_2(g)}$

Only $\mathrm{CO_2}$ is gaseous; $\Delta n = 1 - 0 = 1$

$$K_c = \frac{K_p}{(RT)^1} = \frac{167}{0.0831 \times 1073} = \frac{167}{89.17}$$
$$\boxed{K_c = 1.87}$$

Given: $K_c(\text{forward}) = 6.3 \times 10^{14}$ at $1000\,\mathrm{K}$

Concept: For the reverse reaction, the equilibrium constant is the reciprocal of the forward equilibrium constant.

$$K_c(\text{reverse}) = \frac{1}{K_c(\text{forward})} = \frac{1}{6.3 \times 10^{14}}$$

$$\boxed{K_c(\text{reverse}) = 1.59 \times 10^{-15}}$$

Explanation:

The equilibrium constant expression is written in terms of activities of the species involved. For an ideal solution or gas, the activity is proportional to molar concentration or partial pressure. However, for a pure solid or pure liquid, the activity is defined as unity (1) because their molar concentration (density/molar mass) remains essentially constant throughout the reaction and does not change with the extent of reaction.

For example, the molar concentration of water (pure liquid) is:
$$[\mathrm{H_2O}] = \frac{1000\,\mathrm{g/L}}{18\,\mathrm{g/mol}} \approx 55.5\,\mathrm{mol/L} = \text{constant}$$

Since their concentrations are constant, they are incorporated into the equilibrium constant itself and do not appear explicitly in the $K_c$ expression. Therefore, pure liquids and solids are ignored (their activity = 1) while writing the equilibrium constant expression.

Given:
- Initial moles: $n(\mathrm{N_2}) = 0.482\,\mathrm{mol}$, $n(\mathrm{O_2}) = 0.933\,\mathrm{mol}$, Volume $= 10\,\mathrm{L}$
- $K_c = 2.0 \times 10^{-37}$ (extremely small)

Initial concentrations:
$$[\mathrm{N_2}]_0 = \frac{0.482}{10} = 0.0482\,\mathrm{M}$$
$$[\mathrm{O_2}]_0 = \frac{0.933}{10} = 0.0933\,\mathrm{M}$$

ICE Table: Let $x$ mol/L of $\mathrm{N_2O}$ be formed at equilibrium.

$$2\mathrm{N_2} + \mathrm{O_2} \rightleftharpoons 2\mathrm{N_2O}$$

| | $\mathrm{N_2}$ | $\mathrm{O_2}$ | $\mathrm{N_2O}$ |
|---|---|---|---|
|I| 0.0482 | 0.0933 | 0 |
|C| $-x$ | $-x/2$ | $+x$ |
|E| $0.0482-x$ | $0.0933-x/2$ | $x$ |

$K_c$ expression:
$$K_c = \frac{[\mathrm{N_2O}]^2}{[\mathrm{N_2}]^2[\mathrm{O_2}]} = 2.0 \times 10^{-37}$$

Since $K_c$ is extremely small, $x$ is negligibly small compared to initial concentrations. So:
$$K_c \approx \frac{x^2}{(0.0482)^2(0.0933)}$$
$$x^2 = 2.0 \times 10^{-37} \times (0.0482)^2 \times 0.0933$$
$$x^2 = 2.0 \times 10^{-37} \times 2.323 \times 10^{-3} \times 0.0933$$
$$x^2 = 2.0 \times 10^{-37} \times 2.167 \times 10^{-4}$$
$$x^2 = 4.334 \times 10^{-41}$$
$$x = 6.58 \times 10^{-21}\,\mathrm{mol/L}$$

Equilibrium composition (essentially unchanged):
$$[\mathrm{N_2}] \approx 0.0482\,\mathrm{M} \Rightarrow n(\mathrm{N_2}) = 0.482\,\mathrm{mol}$$
$$[\mathrm{O_2}] \approx 0.0933\,\mathrm{M} \Rightarrow n(\mathrm{O_2}) = 0.933\,\mathrm{mol}$$
$$[\mathrm{N_2O}] = 6.58 \times 10^{-21}\,\mathrm{mol/L} \Rightarrow n(\mathrm{N_2O}) = 6.58 \times 10^{-20}\,\mathrm{mol}$$

The equilibrium mixture essentially contains the same amounts of $\mathrm{N_2}$ and $\mathrm{O_2}$ as initially, with a negligible amount of $\mathrm{N_2O}$.

Given:
- Initial: $n(\mathrm{NO}) = 0.087\,\mathrm{mol}$, $n(\mathrm{Br_2}) = 0.0437\,\mathrm{mol}$, $n(\mathrm{NOBr}) = 0$
- At equilibrium: $n(\mathrm{NOBr}) = 0.0518\,\mathrm{mol}$

From stoichiometry:

For every 2 mol of NOBr formed, 2 mol of NO and 1 mol of $\mathrm{Br_2}$ are consumed.

Moles of NO consumed $= 0.0518\,\mathrm{mol}$ (same as NOBr formed, 2:2 ratio)

Moles of $\mathrm{Br_2}$ consumed $= \dfrac{0.0518}{2} = 0.0259\,\mathrm{mol}$

Equilibrium amounts:
$$n(\mathrm{NO})_{\mathrm{eq}} = 0.087 - 0.0518 = \boxed{0.0352\,\mathrm{mol}}$$
$$n(\mathrm{Br_2})_{\mathrm{eq}} = 0.0437 - 0.0259 = \boxed{0.0178\,\mathrm{mol}}$$

Given: $K_p = 2.0 \times 10^{10}\,\mathrm{bar}^{-1}$, $T = 450\,\mathrm{K}$, $R = 0.0831\,\mathrm{bar\,L\,mol^{-1}\,K^{-1}}$

$\Delta n$ for the reaction:
$$\Delta n = 2 - (2+1) = -1$$

Relation between $K_p$ and $K_c$:
$$K_p = K_c(RT)^{\Delta n}$$
$$K_c = K_p \times (RT)^{-\Delta n} = K_p \times (RT)^{1}$$

$$K_c = 2.0 \times 10^{10} \times (0.0831 \times 450)$$
$$= 2.0 \times 10^{10} \times 37.395$$
$$\boxed{K_c = 7.48 \times 10^{11}\,\mathrm{mol^{-1}\,L}}$$

Given: Initial pressure of HI $= 0.2\,\mathrm{atm}$; equilibrium pressure of HI $= 0.04\,\mathrm{atm}$

Pressure decrease of HI $= 0.2 - 0.04 = 0.16\,\mathrm{atm}$

From stoichiometry: 2 mol HI decomposes to give 1 mol $\mathrm{H_2}$ and 1 mol $\mathrm{I_2}$.

$$p_{\mathrm{H_2}} = p_{\mathrm{I_2}} = \frac{0.16}{2} = 0.08\,\mathrm{atm}$$

$K_p$ expression:
$$K_p = \frac{p_{\mathrm{H_2}} \cdot p_{\mathrm{I_2}}}{(p_{\mathrm{HI}})^2} = \frac{0.08 \times 0.08}{(0.04)^2} = \frac{6.4 \times 10^{-3}}{1.6 \times 10^{-3}}$$

$$\boxed{K_p = 4.0}$$

Note: Since $\Delta n = 0$ for this reaction, $K_c = K_p = 4.0$.

Given: Volume $= 20\,\mathrm{L}$, $K_c = 1.7 \times 10^2$

Concentrations:
$$[\mathrm{N_2}] = \frac{1.57}{20} = 0.0785\,\mathrm{M}$$
$$[\mathrm{H_2}] = \frac{1.92}{20} = 0.096\,\mathrm{M}$$
$$[\mathrm{NH_3}] = \frac{8.13}{20} = 0.4065\,\mathrm{M}$$

Reaction quotient $Q_c$:
$$Q_c = \frac{[\mathrm{NH_3}]^2}{[\mathrm{N_2}][\mathrm{H_2}]^3} = \frac{(0.4065)^2}{(0.0785)(0.096)^3}$$
$$= \frac{0.1652}{0.0785 \times 8.847 \times 10^{-4}} = \frac{0.1652}{6.945 \times 10^{-5}}$$
$$Q_c = 2.38 \times 10^3$$

Comparison:
Q_c = 2.38 \times 10^3 > K_c = 1.7 \times 10^2

Since Q_c > K_c, the reaction is not at equilibrium. The reaction will proceed in the reverse direction (i.e., decomposition of $\mathrm{NH_3}$) to reach equilibrium.

Given: $K_c = \dfrac{[\mathrm{NH_3}]^4[\mathrm{O_2}]^5}{[\mathrm{NO}]^4[\mathrm{H_2O}]^5}$

Interpretation: Products are $\mathrm{NH_3}$ and $\mathrm{O_2}$; reactants are $\mathrm{NO}$ and $\mathrm{H_2O}$.

The balanced chemical equation is:
$$\boxed{4\mathrm{NO}(\mathrm{g}) + 6\mathrm{H_2O}(\mathrm{g}) \rightleftharpoons 4\mathrm{NH_3}(\mathrm{g}) + 5\mathrm{O_2}(\mathrm{g})}$$

Verification: This is the reverse of the catalytic oxidation of ammonia. The $K_c$ expression matches the given expression.

Given: Initial moles: $n(\mathrm{H_2O}) = 1\,\mathrm{mol}$, $n(\mathrm{CO}) = 1\,\mathrm{mol}$; Volume $= 10\,\mathrm{L}$; 40% of water reacts.

Moles reacted: $0.40 \times 1 = 0.4\,\mathrm{mol}$ of $\mathrm{H_2O}$

ICE Table (in moles):

| | $\mathrm{H_2O}$ | $\mathrm{CO}$ | $\mathrm{H_2}$ | $\mathrm{CO_2}$ |
|---|---|---|---|---|
|I| 1 | 1 | 0 | 0 |
|C| $-0.4$ | $-0.4$ | $+0.4$ | $+0.4$ |
|E| 0.6 | 0.6 | 0.4 | 0.4 |

Equilibrium concentrations (in $10\,\mathrm{L}$ vessel):
$$[\mathrm{H_2O}] = \frac{0.6}{10} = 0.06\,\mathrm{M},\quad [\mathrm{CO}] = 0.06\,\mathrm{M}$$
$$[\mathrm{H_2}] = \frac{0.4}{10} = 0.04\,\mathrm{M},\quad [\mathrm{CO_2}] = 0.04\,\mathrm{M}$$

$K_c$:
$$K_c = \frac{[\mathrm{H_2}][\mathrm{CO_2}]}{[\mathrm{H_2O}][\mathrm{CO}]} = \frac{0.04 \times 0.04}{0.06 \times 0.06} = \frac{1.6 \times 10^{-3}}{3.6 \times 10^{-3}}$$
$$\boxed{K_c = 0.444}$$

Given: $K_c = 54.8$ for $\mathrm{H_2 + I_2 \rightleftharpoons 2HI}$; $[\mathrm{HI}]_{\mathrm{eq}} = 0.5\,\mathrm{mol\,L^{-1}}$

Since we started with HI only, the reverse reaction occurs:
$$2\mathrm{HI}(\mathrm{g}) \rightleftharpoons \mathrm{H_2}(\mathrm{g}) + \mathrm{I_2}(\mathrm{g})$$

For this reverse reaction:
$$K_c' = \frac{1}{K_c} = \frac{1}{54.8} = 1.825 \times 10^{-2}$$

Let $[\mathrm{H_2}] = [\mathrm{I_2}] = x$ at equilibrium (by symmetry, since we started with only HI).

$$K_c' = \frac{[\mathrm{H_2}][\mathrm{I_2}]}{[\mathrm{HI}]^2} = \frac{x^2}{(0.5)^2} = 1.825 \times 10^{-2}$$
$$x^2 = 1.825 \times 10^{-2} \times 0.25 = 4.56 \times 10^{-3}$$
$$x = 0.0676\,\mathrm{mol\,L^{-1}}$$

$$\boxed{[\mathrm{H_2}] = [\mathrm{I_2}] = 0.068\,\mathrm{mol\,L^{-1}}}$$

Given: $[\mathrm{ICl}]_0 = 0.78\,\mathrm{M}$, $K_c = 0.14$

ICE Table:

| | $2\mathrm{ICl}$ | $\mathrm{I_2}$ | $\mathrm{Cl_2}$ |
|---|---|---|---|
|I| 0.78 | 0 | 0 |
|C| $-2x$ | $+x$ | $+x$ |
|E| $0.78-2x$ | $x$ | $x$ |

$K_c$ expression:
$$K_c = \frac{[\mathrm{I_2}][\mathrm{Cl_2}]}{[\mathrm{ICl}]^2} = \frac{x \cdot x}{(0.78-2x)^2} = 0.14$$

$$\frac{x^2}{(0.78-2x)^2} = 0.14$$

Taking square root of both sides:
$$\frac{x}{0.78-2x} = \sqrt{0.14} = 0.3742$$

$$x = 0.3742(0.78 - 2x) = 0.2919 - 0.7484x$$
$$x + 0.7484x = 0.2919$$
$$1.7484x = 0.2919$$
$$x = 0.167\,\mathrm{M}$$

Equilibrium concentrations:
$$[\mathrm{ICl}] = 0.78 - 2(0.167) = 0.78 - 0.334 = \boxed{0.446\,\mathrm{M}}$$
$$[\mathrm{I_2}] = [\mathrm{Cl_2}] = \boxed{0.167\,\mathrm{M}}$$

Given: $K_p = 0.04\,\mathrm{atm}$, initial pressure of $\mathrm{C_2H_6} = 4.0\,\mathrm{atm}$, $T = 899\,\mathrm{K}$

ICE Table (in terms of pressure):

| | $\mathrm{C_2H_6}$ | $\mathrm{C_2H_4}$ | $\mathrm{H_2}$ |
|---|---|---|---|
|I| 4.0 | 0 | 0 |
|C| $-p$ | $+p$ | $+p$ |
|E| $4.0-p$ | $p$ | $p$ |

$$K_p = \frac{p_{\mathrm{C_2H_4}} \cdot p_{\mathrm{H_2}}}{p_{\mathrm{C_2H_6}}} = \frac{p^2}{4.0-p} = 0.04$$

$$p^2 = 0.04(4.0-p) = 0.16 - 0.04p$$
$$p^2 + 0.04p - 0.16 = 0$$

Using quadratic formula:
$$p = \frac{-0.04 + \sqrt{(0.04)^2 + 4 \times 0.16}}{2} = \frac{-0.04 + \sqrt{0.0016 + 0.64}}{2}$$
$$= \frac{-0.04 + \sqrt{0.6416}}{2} = \frac{-0.04 + 0.8010}{2} = \frac{0.761}{2} = 0.3805\,\mathrm{atm}$$

Equilibrium pressure of $\mathrm{C_2H_6}$:
$$p_{\mathrm{C_2H_6}} = 4.0 - 0.3805 = 3.619\,\mathrm{atm}$$

Equilibrium concentration (using $pV = nRT$, $c = p/RT$):
$$[\mathrm{C_2H_6}] = \frac{p}{RT} = \frac{3.619}{0.0821 \times 899} = \frac{3.619}{73.81}$$
$$\boxed{[\mathrm{C_2H_6}] = 0.049\,\mathrm{mol\,L^{-1}}}$$

(i) Reaction quotient $Q_c$:
$$Q_c = \frac{[\mathrm{CH_3COOC_2H_5}][\mathrm{H_2O}]}{[\mathrm{CH_3COOH}][\mathrm{C_2H_5OH}]}$$

(ii) Calculation of $K_c$:

Let volume of the system $= V\,\mathrm{L}$.

Initial moles: $n(\mathrm{CH_3COOH}) = 1.00$, $n(\mathrm{C_2H_5OH}) = 0.18$, $n(\mathrm{ester}) = 0$, $n(\mathrm{H_2O}) = 0$

At equilibrium: $n(\mathrm{ester}) = 0.171\,\mathrm{mol}$

Moles consumed: $n(\mathrm{CH_3COOH}) = 0.171$, $n(\mathrm{C_2H_5OH}) = 0.171$

Equilibrium moles:
- $n(\mathrm{CH_3COOH}) = 1.00 - 0.171 = 0.829\,\mathrm{mol}$
- $n(\mathrm{C_2H_5OH}) = 0.18 - 0.171 = 0.009\,\mathrm{mol}$
- $n(\mathrm{ester}) = 0.171\,\mathrm{mol}$
- $n(\mathrm{H_2O}) = 0.171\,\mathrm{mol}$

Since all species are in the same volume $V$, it cancels:
$$K_c = \frac{(0.171/V)(0.171/V)}{(0.829/V)(0.009/V)} = \frac{0.171 \times 0.171}{0.829 \times 0.009}$$
$$= \frac{0.02924}{0.007461} = \boxed{3.92}$$

(iii) Checking if equilibrium is reached:

Initial: $n(\mathrm{CH_3COOH}) = 1.0$, $n(\mathrm{C_2H_5OH}) = 0.5$

After some time: $n(\mathrm{ester}) = 0.214\,\mathrm{mol}$

Moles at this point:
- $n(\mathrm{CH_3COOH}) = 1.0 - 0.214 = 0.786$
- $n(\mathrm{C_2H_5OH}) = 0.5 - 0.214 = 0.286$
- $n(\mathrm{ester}) = 0.214$
- $n(\mathrm{H_2O}) = 0.214$

$$Q_c = \frac{0.214 \times 0.214}{0.786 \times 0.286} = \frac{0.04580}{0.2248} = 0.2036$$

Since $Q_c = 0.204 \neq K_c = 3.92$, equilibrium has not been reached. Since Q_c < K_c, the reaction will proceed in the forward direction.

Given: $[\mathrm{PCl_5}]_{\mathrm{eq}} = 0.05\,\mathrm{mol\,L^{-1}}$, $K_c = 8.3 \times 10^{-2}$

$K_c$ expression:
$$K_c = \frac{[\mathrm{PCl_3}][\mathrm{Cl_2}]}{[\mathrm{PCl_5}]}$$

Since the vessel was initially evacuated and only $\mathrm{PCl_5}$ was introduced, by stoichiometry:
$$[\mathrm{PCl_3}] = [\mathrm{Cl_2}] = x$$

$$8.3 \times 10^{-2} = \frac{x \cdot x}{0.05} = \frac{x^2}{0.05}$$
$$x^2 = 8.3 \times 10^{-2} \times 0.05 = 4.15 \times 10^{-3}$$
$$x = \sqrt{4.15 \times 10^{-3}} = 0.0644\,\mathrm{mol\,L^{-1}}$$

$$\boxed{[\mathrm{PCl_3}] = [\mathrm{Cl_2}] = 6.44 \times 10^{-2}\,\mathrm{mol\,L^{-1}}}$$

Given: $K_p = 0.265$, $p_{\mathrm{CO}}^0 = 1.4\,\mathrm{atm}$, $p_{\mathrm{CO_2}}^0 = 0.80\,\mathrm{atm}$

$K_p$ expression (solids excluded):
$$K_p = \frac{p_{\mathrm{CO_2}}}{p_{\mathrm{CO}}}$$

Check $Q_p$:
Q_p = \frac{0.80}{1.4} = 0.571 > K_p = 0.265

Since Q_p > K_p, the reaction proceeds in the reverse direction.

Let pressure of $\mathrm{CO_2}$ decrease by $x$:

| | $\mathrm{CO}$ | $\mathrm{CO_2}$ |
|---|---|---|
|I| 1.4 | 0.80 |
|C| $+x$ | $-x$ |
|E| $1.4+x$ | $0.80-x$ |

$$K_p = \frac{0.80-x}{1.4+x} = 0.265$$
$$0.80 - x = 0.265(1.4 + x) = 0.371 + 0.265x$$
$$0.80 - 0.371 = x + 0.265x = 1.265x$$
$$0.429 = 1.265x$$
$$x = 0.339\,\mathrm{atm}$$

Equilibrium partial pressures:
$$p_{\mathrm{CO}} = 1.4 + 0.339 = \boxed{1.739\,\mathrm{atm}}$$
$$p_{\mathrm{CO_2}} = 0.80 - 0.339 = \boxed{0.461\,\mathrm{atm}}$$

Given: $K_c = 0.061$; $[\mathrm{N_2}] = 3.0\,\mathrm{M}$, $[\mathrm{H_2}] = 2.0\,\mathrm{M}$, $[\mathrm{NH_3}] = 0.5\,\mathrm{M}$

Reaction quotient $Q_c$:
$$Q_c = \frac{[\mathrm{NH_3}]^2}{[\mathrm{N_2}][\mathrm{H_2}]^3} = \frac{(0.5)^2}{(3.0)(2.0)^3} = \frac{0.25}{3.0 \times 8} = \frac{0.25}{24}$$
$$Q_c = 1.04 \times 10^{-2}$$

Comparison:
Q_c = 1.04 \times 10^{-2} < K_c = 0.061

Since Q_c < K_c, the reaction is not at equilibrium. The reaction will proceed in the forward direction (towards formation of $\mathrm{NH_3}$) to reach equilibrium.

Given: $[\mathrm{BrCl}]_0 = 3.3 \times 10^{-3}\,\mathrm{M}$, $K_c = 32$

ICE Table:

| | $2\mathrm{BrCl}$ | $\mathrm{Br_2}$ | $\mathrm{Cl_2}$ |
|---|---|---|---|
|I| $3.3 \times 10^{-3}$ | 0 | 0 |
|C| $-2x$ | $+x$ | $+x$ |
|E| $3.3\times10^{-3}-2x$ | $x$ | $x$ |

$$K_c = \frac{[\mathrm{Br_2}][\mathrm{Cl_2}]}{[\mathrm{BrCl}]^2} = \frac{x^2}{(3.3\times10^{-3}-2x)^2} = 32$$

Taking square root:
$$\frac{x}{3.3\times10^{-3}-2x} = \sqrt{32} = 5.657$$
$$x = 5.657(3.3\times10^{-3} - 2x) = 1.867\times10^{-2} - 11.314x$$
$$x(1 + 11.314) = 1.867\times10^{-2}$$
$$12.314x = 1.867\times10^{-2}$$
$$x = 1.516\times10^{-3}\,\mathrm{M}$$

Equilibrium concentration of BrCl:
$$[\mathrm{BrCl}] = 3.3\times10^{-3} - 2(1.516\times10^{-3}) = 3.3\times10^{-3} - 3.032\times10^{-3}$$
$$\boxed{[\mathrm{BrCl}] = 2.68\times10^{-4}\,\mathrm{mol\,L^{-1}}}$$