Respuesta :
Answer : The reaction must shift to the product or right to be in equilibrium. The equilibrium concentration of [tex]H_2[/tex] is 7.0 M
Explanation :
Reaction quotient (Qc) : It is defined as the measurement of the relative amounts of products and reactants present during a reaction at a particular time.
First we have to determine the concentration of [tex]CH_4,H_2S,CS_2\text{ and }H_2[/tex].
[tex]\text{Concentration of }CH_4=\frac{\text{Moles of }CH_4}{\text{Volume of solution}}=\frac{1mol}{400mL}\times 1000=5M[/tex]
[tex]\text{Concentration of }H_2S=\frac{\text{Moles of }H_2S}{\text{Volume of solution}}=\frac{2mol}{400mL}\times 1000=2.5M[/tex]
[tex]\text{Concentration of }CS_2=\frac{\text{Moles of }CS_2}{\text{Volume of solution}}=\frac{1mol}{400mL}\times 1000=2.5M[/tex]
[tex]\text{Concentration of }H_2=\frac{\text{Moles of }H_2}{\text{Volume of solution}}=\frac{2mol}{400mL}\times 1000=5M[/tex]
Now we have to determine the value of reaction quotient (Qc).
The given balanced chemical reaction is,
[tex]CH_4(g)+2H_2S(g)\rightarrow CS_2(g)+4H_2(g)[/tex]
The expression for reaction quotient will be :
[tex]Q_c=\frac{[CS_2][H_2]^4}{[CH_4][H_2S]^2}[/tex]
In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.
Now put all the given values in this expression, we get
[tex]Q_c=\frac{(2.5)\times (5)^4}{(2.5)times (5)^2}=25[/tex]
Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.
There are 3 conditions:
When [tex]Q>K[/tex] that means product > reactant. So, the reaction is reactant favored.
When [tex]Q<K[/tex] that means reactant > product. So, the reaction is product favored.
When [tex]Q=K[/tex] that means product = reactant. So, the reaction is in equilibrium.
The given equilibrium constant value is, [tex]K_c=225[/tex]
From the above we conclude that, the [tex]Q<K[/tex] that means reactant > product. So, the reaction is product favored that means reaction must shift to the product or right to be in equilibrium.
Now we have to calculate the concentration of [tex]H_2[/tex] at equilibrium.
The given balanced chemical reaction is,
[tex]CH_4(g)+2H_2S(g)\rightarrow CS_2(g)+4H_2(g)[/tex]
Initial conc. 2.5 5 2.5 5
At eqm. (2.5-x) (5-2x) (2.5+x) (5+4x)
The concentration of [tex]CH_4[/tex] at equilibrium = 2.0 M
As we know that, at equilibrium
(2.5-x) = 2.0 M
x = 0.5 M
The concentration of [tex]H_2[/tex] at equilibrium = (5+4x) = 5 + 4(0.5) = 7.0 M
Therefore, the equilibrium concentration of [tex]H_2[/tex] is 7.0 M