Chemical Equilibrium

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

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

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

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

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

$\left[N{H}_{4}Cl\right]=\frac{2}{60}=\frac{1}{30}M$

$pH=7-\frac{1}{2}P{K}_b-\frac{1}{2}logC$

$=7-\frac{5}{2}-\frac{1}{2}log\left(\frac{1}{30}\right)=5.24$

Ka for C₃H₇COOH = 2 * 10^-5

$pKa=-\mathcal{l}og\left(2*10^{-5}\right)=5-\mathcal{l}og2$  

=5 – 0.3 = 4.7

pH of 0.2 (M) solution = 

$pH=\frac{pKa-\mathcal{l}ogC}{2}$  

$=\frac{1}{2}\left(4.7\right)-\frac{1}{2}\mathcal{l}og\left(0.2\right)$  

$pH=27*10^{-1}$     

Ans 27

K_p = 47.9      T = 288 K

K_C =?                    R = 0.083 L bar/K mol

$N_2{O}_4\left(g\right)?2N{O}_2\left(g\right)$           

Using

$K_p=K_C{\left(RT\right)}^{\Delta ng}$              

Here ;

$\Delta ng=1$                

49.7 = K_C (0.083 * 288)¹

K_C = 2

$\left[C{l}^{-}\right]=10^{-1}M$

$\left[Cr{O_4}^{--}\right]=10^{-3}M$

for AgCl ppt¹, ${\left[A{g}^{+}\right]}_{req}=\frac{1.7*10^{-10}}{10^{-1}}M$  

For Ag₂CrO₄ppt², ${\left[A{g}^{+}\right]}_{req}=\sqrt[]{\frac{1.9*10^{-12}}{10^{-3}}}M$

${\left[A{g}^{+}\right]}_{req}=4.3*10^{-5}M$

Being lower concentration of [Ag⁺] in case of AgCl, it will precipitate first.

Mass of CuSO₄. 5H₂O = 80 g

Volume of solution = 5L

Molar mass of CuSO₄.5H₂O = 249.54g/ml

$Concentration=\frac{moles}{volumeofsolution}$          

$\begin{array}{l}=\frac{0.32}{5}=0.064M\\ =64*10^{-3}M\end{array}$                

50 ml of 1 (M) HCl + 30 ml of 1 (M) NaOH

NaOH    +            HCl  ->          NaCl + H₂O

30 * 1 mmol      50 * 1 mmol     

0 mmol               20 mmol

${\left[H^{+}\right]}_{mix}=\frac{20}{50+30}M=\frac{20}{80}M=\frac{1}{4}M=0.25M$

$x*10^{-4}=6021*10^{-4}$    

$\therefore$ x = 6021

Ans. = 6021

Moles of NH₄HS initially taken = $\frac{5.1}{51}=0.1mol$  

$N{H}_{4}HS\left(S\right)+N{H}_3\left(g\right)+H_{2}S\left(g\right)$                             

t =  $\infty$ 0       0.1 mol               0            0

t =   $\infty$  0.1 (1 - 0.2)   0.1 * 0.2   0.1 * 0.2

$P_{N{H}_3}=\frac{nRT}{V}=\frac{0.1*0.2*0.082*300}{2}=0.246atm=P_{H_{2}S}$

$K_P=P_{N{H}_3}*P_{H_{2}S}={\left(0.246\right)}^2=0.060516=6.05*10^{-2}$

x = 6 (Nearest integer).

Acid = M (OH)2 Salt + H2O

M.E of Acid = me of Base

$\therefore 10*\left(0.1*nf\right)=\left(0.05*2\right)*30$

$\therefore n_f=3$

Thus basicity of acid = 3