Question

Chemistry help please

Answer 1

The reaction isn't yet at equilibrium. The overall reaction will continue to move in the direction of the products.

Assumption: this system is currently at
`\rm 900^(\circ)C`.

Step-by-step explanation:

One way to tell whether a system is at its equilibrium is to compare its reaction quotient
`Q` with the equilibrium constant
`K_c` of the reaction.

The equation for
`Q` is quite similar to that for
`K_c`. The difference between the two is that
`K_c` requires equilibrium concentrations, while
`Q` can be calculated even when the system is on its way to equilibrium.

For this reaction,

`\displaystyle Q = \rm ([CS_2]\cdot [H_2]^(4))/([CH_4]\cdot [H_2S]^(2))`.

Given these concentrations,

`\displaystyle Q = \rm ([CS_2]\cdot [H_2]^(4))/([CH_4]\cdot [H_2S]^(2)) =(1.51* (1.08)^(4))/(1.15* (1.20)^(2)) \approx 1.72`.

The question states that at
`\rm 900^(\circ)C`,
`K_c = 3.59`. Assume that currently this system is also at
`\rm 900^(\circ)C`. (The two temperatures need to be the same since the value of
`K_c` depends on the temperature.)

It turns out that
`Q = K_c`. What does this mean?

• First, the system isn't at equilibrium.
• Second, if there's no external changes, the system will continue to move towards the equilibrium. Temperature might change. However, eventually
  `Q` will be equal to
  `K_c`, and the system will achieve equilibrium.

In which direction will the system move? At this moment,
`Q < K_c`. As time proceeds, the value of
`Q` will increase so that it could become equal to
`K_c`. Recall that
`Q` is fraction.

`\displaystyle Q = \rm ([CS_2]\cdot [H_2]^(4))/([CH_4]\cdot [H_2S]^(2))`

When the value of
`Q` increases, either its numerator becomes larger or its denominator becomes smaller, or both will happen at the same time. However,

• Concentrations on the numerator of
  `Q` are those of the products;
• Concentrations on the denominator of
  `Q` are those of the reactants.

As time proceeds,

• the concentration of the products will increase, while
• the concentration of the reactants will decrease.

In other words, the equilibrium will move towards the products.