Calculate the percent dissociation of H 2 S ( g ) if 0.1 mole of H 2 S is kept in 0.4 litre vessel at 1000 K. For the reaction: 2 H 2 S (g) leftharpoons 2 H 2(g)+ S 2(g) the value of K C is 1.0 × 10-6.

Question

Calculate the percent dissociation of H 2 S ( g ) if 0.1 mole of H 2 S is kept in 0.4 litre vessel at 1000 K. For the reaction: 2 H 2 S (g) leftharpoons 2 H 2(g)+ S 2(g) the value of K C is 1.0 × 10-6. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Molar conc. of H 2 S =(0.1/0.4) mol L-1 =0.25 mol L-1 Suppose degree of dissociation of H 2 S =α then 2 H 2 S leftharpoons 2 H 2+ S 2 0.25 M 0.25(1-α), 0.25 α, (0.25/2) α =0.125 α=0.125 α K C =([ H 2]2[ S 2]/[ H 2 S ]2) 10-6=((0.25 α)2(0.125 α)/[0.25(1-2)]2) Neglecting in comparison to 1 we get 10-6=((0.25 α)2(0.125 α)/(0.25)2)=0.02 % age dissociation =0.02 × 100=2 %

Q. Calculate the percent dissociation of H2​S(g) if 0.1mole of H2​S is kept in 0.4 litre vessel at 1000K. For the reaction : 2H2​S(g)⇌2H2​(g)+S2​(g) the value of KC​ is 1.0×10−6.

Q. Calculate the percent dissociation of $H_{2} S (g)$ if $0.1 m o l e$ of $H_{2} S$ is kept in $0.4$ litre vessel at $1000 K$ .
For the reaction :
$2 H_{2} S (g) ⇌ 2 H_{2} (g) + S_{2} (g)$
the value of $K_{C}$ is $1.0 \times 1 0^{- 6}$ .