Answer:
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 [tex]\rm 900^{\circ}C[/tex].
Explanation:
One way to tell whether a system is at its equilibrium is to compare its reaction quotient [tex]Q[/tex] with the equilibrium constant [tex]K_c[/tex] of the reaction.
The equation for [tex]Q[/tex] is quite similar to that for [tex]K_c[/tex]. The difference between the two is that [tex]K_c[/tex] requires equilibrium concentrations, while [tex]Q[/tex] can be calculated even when the system is on its way to equilibrium.
For this reaction,
[tex]\displaystyle Q = \rm \frac{[CS_2]\cdot [H_2]^{4}}{[CH_4]\cdot [H_2S]^{2}}[/tex].
Given these concentrations,
[tex]\displaystyle Q = \rm \frac{[CS_2]\cdot [H_2]^{4}}{[CH_4]\cdot [H_2S]^{2}} =\frac{1.51\times (1.08)^{4}}{1.15\times (1.20)^{2}} \approx 1.72[/tex].
The question states that at [tex]\rm 900^{\circ}C[/tex], [tex]K_c = 3.59[/tex]. Assume that currently this system is also at [tex]\rm 900^{\circ}C[/tex]. (The two temperatures need to be the same since the value of [tex]K_c[/tex] depends on the temperature.)
It turns out that [tex]Q = K_c[/tex]. 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 [tex]Q[/tex] will be equal to [tex]K_c[/tex], and the system will achieve equilibrium.
In which direction will the system move? At this moment, [tex]Q < K_c[/tex]. As time proceeds, the value of [tex]Q[/tex] will increase so that it could become equal to [tex]K_c[/tex]. Recall that [tex]Q[/tex] is fraction.
[tex]\displaystyle Q = \rm \frac{[CS_2]\cdot [H_2]^{4}}{[CH_4]\cdot [H_2S]^{2}}[/tex]
When the value of [tex]Q[/tex] 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 [tex]Q[/tex] are those of the products;
- Concentrations on the denominator of [tex]Q[/tex] 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.
