A mixture of 2.50mol H2, 1.5 mol CS2, 1.5 mol CH4 and 2mol H2S is placed in a 5.0L reaction vessel. When equilibrium is achieved, the concentration of CH4 has become 0.25M. Whay are the number of moles of CS2 at equilibrium?

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Answer 1 · 2016-10-08T16:03:13

The possible reaction occurring in this case is represented by the following equation

$" "" "2H_2S(g)" "+" "CH_4" "rightleftharpoons" "4H_2(g)" "+" "CS_2(g).....(1)$

The given volume of the reaction mixture is $5L$

The initial no. of moles of $CH_4$ was $1.5mol$

At equilibrium the concentration of $CH_4$ was $0.25M$

So at equilibrium the no. of moles of $CH_4$ will be $0.25xx5=1.25mol$

So the Change in no. of moles of $CH_4$ will be
$=(1.25-1.5)=-0.25mol$

Hence the Change in no. of moles of $CS_2$ will be
$=+0.25mol$

And at equilibrium the no. of moles of $CS_2$ will be

$=(1.5+0.25)mol=1.75mol$

Using stoichiometric ratio in equation (1) the change in other components can be calculated

Associated ICE table shows the no. of moles of different components presents in the reaction mixture at equilibrium

$" "" "2H_2S(g)" "+" "CH_4" "rightleftharpoons" "4H_2(g)" "+" "CS_2(g)$

$I" "" "2" mol " " "" "1.5" mol"" "->" "2.5" mol"" "" "1.5 " mol"$

$C" "-0.5" mol"" "-0.25" mol"->" "+1" mol" " " "+0.25 " mol"$

$E" "" "1.5" mol"" "" "1.25" mol"->" "3.5" mol" " "" "1.75 " mol"$