Question

The equilibrium constant for a gas phase reaction is $Kc=\frac{[CS_{2}][H_{2}]^{4} }{[CH_{4}][H_{2}S]^{2}}$, write the balanced chemical equation corres- ponding to this expression.

Answer 1

"Law of chemical equilibrium":

This law states that at a "given temperature", a "chemical system" may reach a "state" in which a "particular ratio" of "reactant to product" "concentrations" has "constant value".

For the reaction,

$aA\quad +\quad bB\quad \leftrightarrow \quad cC\quad +\quad dD$

$r_{ f }\quad =\quad k_{ f }[A]^{ a }[B]^{ b }$

$r_{ b }\quad =\quad k_{ b }[C]^{ c }[D]^{ d }$

At equilibrium

$r_{ f }\quad =\quad r_{ b }$

${ K }_{ eq }\quad =\quad \frac { { K }_{ f } }{ { K }_{ b } } =\quad \left( \frac { [C]^{ c }[D]^{ d } }{ [A]^{ a }[B]^{ b } }  \right)$

Therefore, the given expression is ${ K }_{ c }\quad =\quad \frac { [CS_{ 2 }][H_{ 2 }]^{ 4 } }{ [CH_{ 4 }][H_{ 2 }S]^{ 2 } }$

Let's write the chemical reaction as follows

${ CH }_{ 4 }(g)\quad \quad +\quad 2{ H }_{ 2 }S(g)\quad \leftrightarrow \quad C{ S }_{ 2 }(g)\quad +\quad 4{ H }_{ 2 }(g)$"

Answer 2

"Law of chemical equilibrium":

This law states that at a "given temperature", a "chemical system" may reach a "state" in which a "particular ratio" of "reactant to product" "concentrations" has "constant value".

For the reaction,

aA\quad +\quad bB\quad \leftrightarrow \quad cC\quad +\quad dDaA+bB↔cC+dD

r_{ f }\quad =\quad k_{ f }[A]^{ a }[B]^{ b }r

f

=k

f

[A]

a

[B]

b

r_{ b }\quad =\quad k_{ b }[C]^{ c }[D]^{ d }r

b

=k

b

[C]

c

[D]

d

At equilibrium

r_{ f }\quad =\quad r_{ b }r

f

=r

b

Therefore, the given expression is { K }_{ c }\quad =\quad \frac { [CS_{ 2 }][H_{ 2 }]^{ 4 } }{ [CH_{ 4 }][H_{ 2 }S]^{ 2 } }K

c

=

[CH

4

][H

2

S]

2

[CS

2

][H

2

]

4

Let's write the chemical reaction as follows

{ CH }_{ 4 }(g)\quad \quad +\quad 2{ H }_{ 2 }S(g)\quad \leftrightarrow \quad C{ S }_{ 2 }(g)\quad +\quad 4{ H }_{ 2 }(g)CH

4

(g)+2H

2

S(g)↔CS

2

(g)+4H

2

(g) "