Equilibrium

$$\begin{gathered} {\mathrm{CO}}_2\left(\mathrm{g}\right) + \mathrm{C}\left(\mathrm{s}\right) \rightleftharpoons 2\mathrm{CO}\left(\mathrm{g}\right) \\ \mathrm{At} \mathrm{t}=0 {\mathrm{p}}_0 - - \\ \mathrm{At} \mathrm{t}=\mathrm{teq} {\mathrm{p}}_0-\mathrm{p} 2\mathrm{p} \\ {\mathrm{P}}_{\mathrm{T}} ={\mathrm{P}}_0+\mathrm{P} \Rightarrow \mathrm{p} = (0.8-0.6) =0.2, {\mathrm{p}}_{\mathrm{CO}} =0.4 \mathrm{and} {\mathrm{p}}_{{\mathrm{CO}}_2}=0.4 \\ {\mathrm{K}}_{\mathrm{p}} =\frac{{\mathrm{p}}_{{\left(\mathrm{CO}\right)}^2}}{{\mathrm{p}}_{{\mathrm{CO}}_2}}=\frac{{(0.4)}^2}{0.4}=0.4 \end{gathered}$$.

Le-Chatelier’s principle applies to equilibrium involving gases or solutions. It is not applicable to Fe(s) + S(s) ⇌ FeS(s)as solids do not affect equilibrium. Correct answer: (B).

Strong acid HCl fully ionizes, giving the lowest pH. Weak acids HCOOH and CH₃COOH partially ionize, with CH₃COOH being weaker. Order: HCl < HCOOH < CH₃COOH. Correct answer: (C).

$$\begin{gathered} {\left({\mathrm{Zr}}^{4+}\right)}_3{\left(P{O_4}^{-3}\right)}_4 \to 3{\left(\mathrm{Zr}\right)}^{4+} +4(P{O_4)}^{-3} \\ 3\mathrm{s} 4\mathrm{s} \end{gathered}$$

$$\begin{gathered} {\mathrm{K}}_{\mathrm{sp}}={\left(3\mathrm{s}\right)}^3{\left(4\mathrm{s}\right)}^4 = 6912 {\mathrm{s}}^7 \\ \mathrm{s}={\left(\frac{{\mathrm{K}}_{\mathrm{sp}}}{ 6912}\right)}^{\frac{1}{7}} \end{gathered}$$

$$\begin{gathered} Z{r}_3{\left(P{O}_4\right)}_4 \left(s\right) \rightleftharpoons 3{\mathrm{Zr}}^{4+}\left(\mathrm{aq}\right) +4{{\mathrm{PO}}_4}^{3-}\left(\mathrm{aq}\right) \\ 3\mathrm{s} 3\mathrm{s} \\ {\mathrm{K}}_{\mathrm{sp}} = {\left(3\mathrm{s}\right)}^3{\left(4\mathrm{s}\right)}^{4 }=6912 {\mathrm{s}}^7 \\ \mathrm{s} ={\left(\frac{{\mathrm{K}}_{\mathrm{sp}}}{6912}\right)}^{1/7} \end{gathered}$$

$$\begin{gathered} {\mathrm{CO}}_2\left(\mathrm{g}\right) + \mathrm{C}\left(\mathrm{s}\right) \rightleftharpoons 2\mathrm{CO}\left(\mathrm{g}\right) \\ \mathrm{At} \mathrm{t}=0 {\mathrm{p}}_0 - - \\ \mathrm{At} \mathrm{t}=\mathrm{teq} {\mathrm{p}}_0-\mathrm{p} 2\mathrm{p} \\ {\mathrm{P}}_{\mathrm{T}} ={\mathrm{P}}_0+\mathrm{P} \Rightarrow \mathrm{p} = (0.8-0.6) =0.2, {\mathrm{p}}_{\mathrm{CO}} =0.4 \mathrm{and} {\mathrm{p}}_{{\mathrm{CO}}_2}=0.4 \\ {\mathrm{K}}_{\mathrm{p}} =\frac{{\mathrm{p}}_{{\left(\mathrm{CO}\right)}^2}}{{\mathrm{p}}_{{\mathrm{CO}}_2}}=\frac{{(0.4)}^2}{0.4}=0.4 \end{gathered}$$.

$N_2{O}_5\left(g\right)\to 2N{O}_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)$

Let the initial pressure of $N_2{O}_5$ be $P_0$.
Initial moles: Only $N_2{O}_5$ is present $\to n_0$ $=P_0$ (as per ideal gas law).
Let x be the fraction of reaction completed.
For 1 mole of $N_2{O}_5$ decomposing:
2 moles of $N{O}_2$ are formed.
0.5 moles of $O_2$ are formed.

Given that 50 % of the reaction is completed (x =0.5):
$N_2{O}_5$ decomposed: $0.5P_0$
$N{O}_2$ formed: $2\times 0.5P_0=P_0$
$O_2$ formed: $0.5\times 0.5P_0=0.25P_0$
Remaining $N_2{O}_5:P_0-0.5P_0=0.5P_0$
$P_{\text{final }}=P_{N_2{O}_5}+P_{N{O}_2}+P_{O_2}$
$P_{\text{final }}=0.5P_0+P_0+0.25P_0=1.75P_0=\frac{7}{4}{P}_0$

$K_p = K_c$ only when the total moles of gaseous products = total moles of gaseous reactants $(\Delta n_g = 0).$

• $2\,\mathrm{BrCl(g)\,\to\,Br_2(g) + Cl_2(g):}$ 2 → 2 (Δn = 0), so $K_p=K_c.$
• $\mathrm{PCl_5(g)\,\to\,PCl_3(g) + Cl_2(g):}$ 1 → 2 (Δn = +1), so $K_p\neq K_c.$
• $\mathrm{H_2(g) + I_2(g)\,\to\,2HI(g):}$ 2 → 2 (Δn = 0), so $K_p=K_c.$
• $\mathrm{CO(g) + H_2O(g)\,\to\,CO_2(g) + H_2(g):}$ 2 → 2 (Δn = 0), so $K_p=K_c.$

First, determine the equilibrium constant from the given data:

$K_c\text{\hspace{0.17em}=\hspace{0.17em}}\frac{\left[{\text{N}}_{\text{2}}\right]\left[{\text{O}}_{\text{2}}\right]}{{\left[\text{NO}\right]}^{\text{2}}}\text{=}\frac{(\text{3}.\text{0}\times {\text{10}}^{\text{-3}})(\text{4}.\text{2}\times {\text{10}}^{\text{-3}})}{{(\text{2}.\text{8}\times {\text{10}}^{\text{-3}})}^{\text{2}}}\approx \text{1}.\text{61}.$

Next, start with 0.1 M NO and let $2z$ be the amount of NO that dissociates:

$$[NO]_{\text{eq}} = 0.1 - 2z,\quad [N_2]_{\text{eq}} = z,\quad [O_2]_{\text{eq}} = z.$$

Then

$$K_c=1.61 = \frac{z \times z}{(0.1 - 2z)^2} = \frac{z^2}{(0.1 - 2z)^2}.$$

Solving yields $z \approx 0.036$.

Degree of dissociation $\alpha$ is the fraction of the initial NO reacting:

$\alpha =\frac{\text{moles\hspace{0.17em}of\hspace{0.17em}NOused}}{\text{initial\hspace{0.17em}moles}}=\frac{2z}{0.1}=\frac{2\times 0.036}{0.1}=0.72\approx \text{0}.\text{717}.$

\(\begin{aligned}
& A_2 B(g) \rightleftharpoons A_2(g)+\frac{1}{2} B_2(g) \\
& t=0 ~p_0 \\
& t=t_{\text {eq }} ~p_0(1-\alpha) \quad p_0 \alpha \quad p_0 \frac{\alpha}{2} \\
& P=p_0+p_0 \frac{\alpha}{2} \\
& P=p_0\left(1+\frac{\alpha}{2}\right) \quad\left(P \approx p_0\right)
\end{aligned}\)

\(\begin{aligned}
& \text { At equilibrium } K_p=\frac{\left(p_{A_2}\right)\left(p_{B_2}\right)}{\left(p_{A_2 B}\right)}=(\alpha \ll 1) \\
& k_p=\frac{\left(p_0 \alpha\right)\left(p_0 \frac{\alpha}{2}\right)^{\frac{1}{2}}}{p_0(1-\alpha)}=k_p=\alpha\left(p \frac{\alpha}{2}\right)^{\frac{1}{2}} \\
& \frac{K_p}{p^{\frac{1}{2}}}=\frac{\alpha^{3 / 2}}{2^{1 / 2}} \\
& \frac{2 K_p^2}{P}=\alpha^3 \\
& \sqrt[3]{\frac{2 K_p^2}{P}}=\alpha
\end{aligned}\)

$$\begin{gathered} {\mathrm{CO}}_2\left(\mathrm{g}\right) + \mathrm{C}\left(\mathrm{s}\right) \rightleftharpoons 2\mathrm{CO} \left(\mathrm{g}\right) \\ 0.5 - \\ 0.5 - \mathrm{x} 2\mathrm{x} \\ {\mathrm{P}}_{\mathrm{T}}= 0.5 + \mathrm{x}= 0.8 \\ \mathrm{x} = 0.8- 0.5 = 0.3 \\ {\mathrm{K}}_{\mathrm{P}} = \frac{{(0.6)}^2}{0.2} = 1.8 \end{gathered}$$

\(\begin{aligned}
& A_2 B(g) \rightleftharpoons A_2(g)+\frac{1}{2} B_2(g) \\
& t=0 ~p_0 \\
& t=t_{\text {eq }} ~p_0(1-\alpha) \quad p_0 \alpha \quad p_0 \frac{\alpha}{2} \\
& P=p_0+p_0 \frac{\alpha}{2} \\
& P=p_0\left(1+\frac{\alpha}{2}\right) \quad\left(P \approx p_0\right)
\end{aligned}\)

\(\begin{aligned}
& \text { At equilibrium } K_p=\frac{\left(p_{A_2}\right)\left(p_{B_2}\right)}{\left(p_{A_2 B}\right)}=(\alpha \ll 1) \\
& k_p=\frac{\left(p_0 \alpha\right)\left(p_0 \frac{\alpha}{2}\right)^{\frac{1}{2}}}{p_0(1-\alpha)}=k_p=\alpha\left(p \frac{\alpha}{2}\right)^{\frac{1}{2}} \\
& \frac{K_p}{p^{\frac{1}{2}}}=\frac{\alpha^{3 / 2}}{2^{1 / 2}} \\
& \frac{2 K_p^2}{P}=\alpha^3 \\
& \sqrt[3]{\frac{2 K_p^2}{P}}=\alpha
\end{aligned}\)

$\mathrm{Q}=\frac{\left[\mathrm{B}\right]\left[\mathrm{C}\right]}{{\left[\mathrm{A}\right]}^2}$

$\mathrm{Q}=\frac{\left(2\times {10}^{-3}\right)\left(2\times {10}^{-3}\right)}{{\left(2\times {10}^{-3}\right)}^2}$

$\mathrm{Q}=\frac{4\times {10}^{-6}}{4\times {10}^{-6}}$

Q = 1

as Q_C > K_C so the reaction will proceed in the reverse direction.

We know the dissociation equation for a weak acid HA :

$K_a=\frac{\left[H^{+}\right]\left[A^{-}\right]}{\left[HA\right]}$

Since $\left[H^{+}\right]=Cx$ and $\left[A^{-}\right]=Cx$, while [HA] $=C(1-x)$, we get:

$K_a=\frac{C{x}^2}{1-x}$

Taking $\log$ on both sides:

$\log K_a=\log C+2\log x-\log (1-x)$

Since $pH=-\log \left[H^{+}\right]$, we get:

$pH=-\log C-\log x$

Similarly, $p{K}_a=-\log K_{\alpha }$, so:

$p{K}_a=-\log C-2\log x+\log (1-x)$

Now, subtracting:

$$\begin{gathered} pH-p{K}_a=(-\log C-\log x)-(-\log C- \\ 2\log x+\log (1-x\left)\right)=\log x-\log (1-x) \\ =\log \frac{x}{1-x} \end{gathered}$$

• When a reaction is reversed, the new equilibrium constant is the inverse of the original.
• When a reaction is multiplied by a factor n, the new equilibrium constant is the original raised to the power of n.

Since pressure is doubled at constant temperature, the concentration of gases increases proportionally (PV = nRT). So, both reactants and products increase in concentration.
Since fewer moles of gas are on the right side (products), the equilibrium will shift forward to reduce the total pressure.

$\mathrm{Ice} \rightleftharpoons \mathrm{water}$

According to le Chateliar principle, On increasing pressure, the equilibrium shifts towards the substances having more density. The density of water is more than ice , then the reaction procced in a forward reaction.

As water is heated to 80°C, the auto-ionization of water (${\mathrm{K}}_{\mathrm{w}}$) increases, leading to a higher H⁺ concentration. Since pH = -log[H⁺], an increase in H⁺ results in a lower pH, though water remains neutral as [H⁺] = [OH⁻].

$$\begin{gathered} \mathrm{HA} \left(\mathrm{aq}\right) \rightleftharpoons {\mathrm{H}}^{+}\left(\mathrm{aq}\right) +{\mathrm{A}}^{-}\left(\mathrm{aq}\right) \\ \mathrm{C} - - \\ \mathrm{C}-\mathrm{Cx} \mathrm{Cx} \mathrm{Cx} \\ {\mathrm{K}}_{\mathrm{a}} =\frac{\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{A}}^{-}\right]}{\mathrm{HA}}=\frac{\mathrm{C\alpha }\times \mathrm{C\alpha }}{\mathrm{C}(1-\mathrm{x})} \\ \left[{\mathrm{H}}^{+}\right] =\mathrm{C\alpha } =\frac{{\mathrm{K}}_{\mathrm{a}}(1-\mathrm{x})}{\mathrm{x}} \\ \mathrm{pH}= -\log \left[{\mathrm{H}}^{+}\right] =-\log \frac{\left[{\mathrm{K}}_{\mathrm{a}}(1-\mathrm{x})\right]}{\mathrm{x}} \\ \mathrm{pH} ={\mathrm{PK}}_{\mathrm{a}} -\log \frac{(1-\mathrm{x})}{\mathrm{x}} \\ \mathrm{pH}-{\mathrm{PK}}_{\mathrm{a}} =\log \frac{\mathrm{x}}{1-\mathrm{x}} \end{gathered}$$

$\mathrm{AgCl}\left(\mathrm{s}\right) \leftrightharpoons {\mathrm{Ag}}^{+}\left(\mathrm{aq}\right) + {\mathrm{Cl}}^{–}\left(\mathrm{aq}\right)$

$\left[{\mathrm{Ag}}^{+}\right] = \mathrm{s} \mathrm{and} \left[{\mathrm{Cl}}^{–}\right]=\mathrm{s}$

$\mathrm{K}\mathrm{sp}=\left[{\mathrm{Ag}}^{+}\right]\left[{\mathrm{Cl}}^{–}\right]={\mathrm{s}}^2$

$s=\sqrt{K_{sp}}=\sqrt{{10}^{-10}}={10}^{-5}M$

${\mathrm{K}}_{\mathrm{sp}}=\left[{\mathrm{Ag}}^{+}\right]\left[{\mathrm{Cl}}^{-}\right]={\mathrm{s}}^{\text{'}}({\mathrm{s}}^{\text{'}}+0.1)$

$\begin{array}{l}{\mathrm{K}}_{\mathrm{sp}}={\mathrm{s}}^{\text{'}}(0.1)\\ {10}^{-10}={\mathrm{s}}^{\text{'}}(0.1)\\ {\mathrm{s}}^{\text{'}}=\frac{{10}^{-10}}{0.1}={10}^{-9}\mathrm{M}\end{array}$

$\mathrm{Ratio}=\frac{\mathrm{Solubility} \mathrm{in} \mathrm{KCl}}{\mathrm{Solubility} \mathrm{in} \mathrm{water}}=\frac{\mathrm{s}\text{'}}{\mathrm{s}}=\frac{{10}^{-9}}{{10}^{-5}}={10}^{-4}$

$$\begin{gathered} {\mathrm{X}}_2\mathrm{Y}\left(\mathrm{s}\right) \rightleftharpoons {\mathrm{X}}_2\left(\mathrm{g}\right) + \frac{1}{2}{\mathrm{Y}}_2\left(\mathrm{g}\right) \\ .... \mathrm{P}\text{'} \raisebox{1ex}{$\mathrm{P}\text{'}$}\!\left/ \!\raisebox{-1ex}{$2$}\right. \\ {\mathrm{P}}_{\mathrm{T}} =\frac{3\mathrm{P}\text{'}}{2}\Rightarrow \mathrm{P}\text{'}=\frac{2{\mathrm{P}}_{\mathrm{T}}}{3} \\ {\mathrm{K}}_{\mathrm{p}} = \left({\mathrm{P}}_{{\mathrm{X}}_2}\right){\left({\mathrm{P}}_{{\mathrm{X}}_2}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \\ {\mathrm{K}}_{\mathrm{p}} =\mathrm{P}\text{'}{\left(\frac{\mathrm{P}\text{'}}{2}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} = \frac{{\left(\mathrm{P}\text{'}\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{\sqrt{2}} =\frac{{\left({\displaystyle \frac{2\mathrm{P}}{3}}\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{\sqrt{2}}=\frac{2{\mathrm{P}}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{3^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}} \end{gathered}$$

The given compound is $Cr(OH{)}_3$, which dissociates in water as:

$Cr(OH{)}_3\rightleftharpoons C{r}^{3+}+3O{H}^{-}$

The solubility product constant ( $K_{\text{sp }}$ ) expression for $Cr(OH{)}_3$ is:

$K_{sp}=\left[C{r}^{3+}\right]{\left[O{H}^{-}\right]}^3$

If S is the molar solubility of $Cr(OH{)}_3$, then:

$\left[C{r}^{3+}\right]=S,\left[O{H}^{-}\right]=3S$

Substituting these in the above expression:

$$\begin{gathered} K_{sp} =S\times 27 S^3 \\ {K}_{sp} =27S^4 \end{gathered}$$

Solving for S :

$S=\sqrt[4]{\frac{K_{sp}}{27}}$

Given $K_{sp}=1.6\times {10}^{-30}$, we get:

$S=\sqrt[4]{\frac{1.6\times {10}^{-30}}{27}}$

Ice (solid) has a greater volume than water (liquid). When pressure is increased, the equilibrium will shift towards the phase with the lower volume to minimize the effect of the pressure increase.

Since water has a lower volume than ice, the equilibrium will shift towards the formation of more water.

Compounds with higher $K_{sp}$ values are more soluble in water, while compounds with lower $K_{sp}$ values are less soluble.
Based on the $K_{sp}$ values, the increasing order of solubility product is:

$HgS<PbS<AgBr<Ca(OH{)}_2.$

$$\begin{gathered} A{B}_2\left(g\right)\rightleftharpoons AB\left(g\right)+\frac{1}{2}{B}_2\left(g\right) \\ t_{\text{eq. }}=\frac{(1-x)}{1+\frac{x}{2}}p·\frac{xp}{1+\frac{x}{2}}·\frac{\frac{x}{2}p}{1+\frac{x}{2}} \end{gathered}$$
Since $x\ll 1$, the approximations $1+\frac{x}{2}\approx 1$ and $1-x\approx 1$ hold, simplifying the expression. Rewriting the equilibrium constant expression:

$$\begin{gathered} K_p=\frac{\left(xp\right)·{\left(\frac{xp}{2}\right)}^{\frac{1}{2}}}{p} \\ K_p^2=x^2·\frac{xp}{2} \\ x=\sqrt[3]{\frac{2K_p^2}{p}} \end{gathered}$$

As water is heated degree of ionisation increases
[H+] ion concentration increases pH decreases.