Question

Look at sample problem 17.10 in the 8th edition Silberberg book.

Answer 1

Answer : The reaction must shift to the product or right to be in equilibrium.  The equilibrium concentration of $H_2$ 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 $CH_4,H_2S,CS_2\text{ and }H_2$.

$\text{Concentration of }CH_4=\frac{\text{Moles of }CH_4}{\text{Volume of solution}}=\frac{1mol}{400mL}\times 1000=5M$

$\text{Concentration of }H_2S=\frac{\text{Moles of }H_2S}{\text{Volume of solution}}=\frac{2mol}{400mL}\times 1000=2.5M$

$\text{Concentration of }CS_2=\frac{\text{Moles of }CS_2}{\text{Volume of solution}}=\frac{1mol}{400mL}\times 1000=2.5M$

$\text{Concentration of }H_2=\frac{\text{Moles of }H_2}{\text{Volume of solution}}=\frac{2mol}{400mL}\times 1000=5M$

Now we have to determine the value of reaction quotient (Qc).

The given balanced chemical reaction is,

$CH_4(g)+2H_2S(g)\rightarrow CS_2(g)+4H_2(g)$

The expression for reaction quotient will be :

$Q_c=\frac{[CS_2][H_2]^4}{[CH_4][H_2S]^2}$

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

$Q_c=\frac{(2.5)\times (5)^4}{(2.5)times (5)^2}=25$

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 $Q>K$ that means product > reactant. So, the reaction is reactant favored.

When $Q$ that means reactant > product. So, the reaction is product favored.

When $Q=K$ that means product = reactant. So, the reaction is in equilibrium.

The given equilibrium constant value is, $K_c=225$

From the above we conclude that, the $Q$ 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 $H_2$ at equilibrium.

The given balanced chemical reaction is,

                     $CH_4(g)+2H_2S(g)\rightarrow CS_2(g)+4H_2(g)$

Initial conc.   2.5            5               2.5          5

At eqm.       (2.5-x)       (5-2x)       (2.5+x)    (5+4x)

The concentration of $CH_4$ at equilibrium = 2.0 M

As we know that, at equilibrium

(2.5-x) = 2.0 M

x = 0.5 M

The concentration of $H_2$ at equilibrium = (5+4x) = 5 + 4(0.5) = 7.0 M

Therefore, the equilibrium concentration of $H_2$ is 7.0 M