Equilibrium

16.00
X + Y? 2Z
Initial 1.0 1.5 0.5
At equ m 1-x 1.5-x 0.5+2x
equ m conc 0.75 1.25 1.0
0.5 + 2x = 1
x = 0.25
K = 1 / (0.75 * 1.25)
K = 16/15
x = 16.00

$500\mathrm{m}\mathrm{L}$ has $\mathrm{H}\mathrm{C}\mathrm{l}=25*{10}^{-3}\mathrm{m}\mathrm{o}\mathrm{l}=25\mathrm{m}\cdot \mathrm{m}\mathrm{o}\mathrm{l}$

$120\mathrm{m}\mathrm{L}$ has $=1\mathrm{m}\cdot \mathrm{m}\mathrm{o}\mathrm{l}$

and $500\mathrm{m}\mathrm{L}$ has ${\mathrm{C}\mathrm{H}}_3\mathrm{C}\mathrm{O}\mathrm{O}\mathrm{H}=\frac{1}{20}*{10}^3\mathrm{m}\cdot \mathrm{m}\mathrm{o}\mathrm{l}$

so $20\mathrm{m}\mathrm{L}$ has $\frac{{10}^3}{20}*\frac{20}{500}=2\mathrm{m}\cdot \mathrm{m}\mathrm{o}\mathrm{l}$

$\mathrm{N}\mathrm{a}\mathrm{O}\mathrm{H}$ added in $20\mathrm{m}\mathrm{L}$ is $5*\frac{1}{2}=2.5\mathrm{m}\cdot \mathrm{m}\mathrm{o}\mathrm{l}$

So, $\mathrm{N}\mathrm{a}\mathrm{O}\mathrm{H}$ (left) $+{\mathrm{C}\mathrm{H}}_3\mathrm{C}\mathrm{O}\mathrm{O}\mathrm{H}?{\mathrm{C}\mathrm{H}}_3\mathrm{C}\mathrm{O}\mathrm{O}\mathrm{H}+{\mathrm{H}}_2\mathrm{O}$
1.5
2
1.5

$\Rightarrow \mathrm{p}\mathrm{H}={\mathrm{p}}^{{\mathrm{K}}_{\mathrm{a}}}+\mathrm{l}\mathrm{o}\mathrm{g}?\frac{\text{salt}}{\text{acid}}=4.75+\mathrm{l}\mathrm{o}\mathrm{g}?0.4771=5.2271$

A + B? 2C
Initial: 1, 1
At eq: 1-x, 1+2x
K = [C]²/ ( [A] [B]) = (1+2x)²/ (1-x)² = 100
(1+2x)/ (1-x) = 10
1+2x = 10-10x => 12x = 9 => x = 3/4
[C] = 1+2x = 1+2 (3/4) = 1+1.5 = 2.5M = 25 * 10? ¹M

PCl? (g)? PCl? (g) + Cl? (g); Kc = 1.844
t=0: 3, 0, 0
equilibrium: 3-x, x
Kc = x² / (3-x) = 1.844
x² + 1.844x - 5.532 = 0
x = (-1.844 + √ (1.844² - 4 (1) (-5.532)/2 = (-1.844 + √25.528)/2 ≈ 1.604
At equilibrium number of moles of PCl? = (3 - 1.604) = 1.396 mol
= 1396 * 10? ³ mol

pH = $\frac{1}{2}\left[p{K}_w+p{K}_a-p{K}_b\right]$ $\frac{}{}{}_{}{}_{}{}_{}$

$=\frac{1}{2}\left[14+4.75-5.23\right]$

$=6.76\cong 7$

A (g) -> B (g)        K_P = 100

$\Delta G$ at 300 K and 1 atm

Using

$\Delta G=-RTlnK{}_P$           

$\Delta G=-R*300ln100$            

=-R * 300 * 2 * 2.3

$\Delta G=-1380R$            

So; x = 1380.

$\underset{SS}{AgC{N}_{\left(S\right)}?A{g^{+}}_{\left(aq\right)}+C{N^{-}}_{\left(aq\right)}}$            

$C{N}^{-}+H^{+}?HCN$

Before reaction     S            10^-3       0

After reaction       0            10^-3       S

$\therefore HCN?H^{+}+C{N}^{-}$

$S^2=\frac{2.2*10^{-16}}{6.2*10^{-7}}=\frac{2.2}{6.2}*10^{-9}$

$S=\sqrt[]{\frac{2.2}{6.2}*10^{-9}}=\sqrt[]{\frac{22}{6.2}*10^{-10}}$

$=\sqrt[]{3.54*10^{-10}}=1.88*10^{-5}=1.9*10^{-5}$           

$2?+C,K_C=4*10^{-3}$

At a given time $t,Q_c$ is to be calculated and been compared with $K_c$ .

$Q_C=\frac{\left[B\right]\left[C\right]}{{[A]}^2}=\frac{\left(2*10^{-3}\right)\left(2*10^{-3}\right)}{{\left(2*10^{-3}\right)}^2}$

$Q_C=1$