Equilibrium Processes

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

Mass of CuSO₄. 5H₂O = 80 g

Volume of solution = 5L

Molar mass of CuSO₄.5H₂O = 249.54g/ml

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

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

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).

For equation A + B $?C+D$  

  $K_C=\frac{\left[C\right]\left[D\right]}{\left[A\right]\left[B\right]}=100at298K$             

Here;    A +  B  $?$ C    +    D

t = 0       1M        1M

eq.         (1-x)      (1-x)      (1-x)      (1-x)

$Kc=\frac{\left(1+x\right)*\left(1+x\right)}{\left(1-x\right)*\left(1-x\right)}$

$\therefore x=\frac{9}{11}$  

At equilibrium, concentration of D is = 1 + $\frac{9}{11}=\frac{11+9}{11}=\frac{20}{11}$  

= 1.818 = 181.8 * 10^-2 M

Ans. = 182 (the nearest integer)