1. Tardigrade
  2. Question
  3. Chemistry
  4. The equilibrium constant for the reaction between CH 4( g ) and H 2 S ( g ) to form CS 2( g ) and H 2( g ), at 1173 K is 3.6 . For the following composition of the reaction mixture, [ CH 4]=1.0 M ,[ H 2 S ]=1.20 M [ CS 2]=2.0 M ,[ H 2]=1.8 M decide which of the following option is correct? [ CH 4( g )+2 H 2 S ( g ) leftharpoons CS 2( g )+4 H 2( g )]

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Q. The equilibrium constant for the reaction between $CH _{4}( g )$ and $H _{2} S ( g )$ to form $CS _{2}( g )$ and $H _{2}( g )$, at $1173 \,K$ is $3.6 .$ For the following composition of the reaction mixture,
$\left[ CH _{4}\right]=1.0 M ,\left[ H _{2} S \right]=1.20\, M$
$\left[ CS _{2}\right]=2.0 M ,\left[ H _{2}\right]=1.8\, M$
decide which of the following option is correct?
$\left[ CH _{4}( g )+2 H _{2} S ( g ) \rightleftharpoons CS _{2}( g )+4 H _{2}( g )\right]$

Solution:

$k _{ c }=3.6$ (given)
$CH _{4( g )}+2 H _{2} S \rightleftharpoons CS _{2( g )}+4 H _{2( g )}$
$Q _{ c }=\frac{\left[ H _{2}\right]^{4}\left[ CS _{2}\right]}{\left[ H _{2} S \right]^{2}\left[ CH _{4}\right]}$
$Q _{ c }=\frac{(1.8)^{4}}{(1.2)^{2}} \cdot 2=14.5 g$
$Q _{ c }> k _{c}$
Reaction will shift to form more $H _{2} S$