Respuesta :
Answer:
I. Changing the pressure:
Increasing the pressure: the amount of Hâ‚‚S(g) will increase.
Decreasing the pressure: the amount of Hâ‚‚S(g) will decrease.
II. Changing the temperature:
Increasing the temperature: the amount of Hâ‚‚S(g) will decrease.
Decreasing the temperature: the amount of Hâ‚‚S(g) will increase.
III. Changing the Hâ‚‚ concentration:
Increasing the Hâ‚‚ concentration: the amount of Hâ‚‚S(g) will increase.
Decreasing the Hâ‚‚ concentration: the amount of Hâ‚‚S(g) will decrease.
Explanation:
Le Châtelier's principle states that when there is an dynamic equilibrium, and this equilibrium is disturbed by an external factor, the equilibrium will be shifted in the direction that can cancel the effect of the external factor to reattain the equilibrium.
I. Changing the pressure:
When there is an increase in pressure, the equilibrium will shift towards the side with fewer moles of gas of the reaction. And when there is a decrease in pressure, the equilibrium will shift towards the side with more moles of gas of the reaction.
For the reaction: CH₄(g) + 2H₂S(g) ⇄ CS₂(g) + 4H₂(g),
The reactants side (left) has 3.0 moles of gases and the products side (right) has 5.0 moles of gases.
Increasing the pressure: will shift the reaction to the side with lower moles of gas (left side), amount of Hâ‚‚S(g) will increase.
Decreasing the pressure: will shift the reaction to the side with lower moles of gas (right side), amount of Hâ‚‚S(g) will decrease.
II. Changing the temperature
The reaction is endothermic since the sign of ΔH is positive.
So the reaction can be represented as:
CH₄(g) + 2H₂S(g) + heat ⇄ CS₂(g) + 4H₂(g).
Increasing the temperature:
The T is a part of the reactants, increasing the T increases the amount of the reactants. So, the reaction will be shifted to the right to suppress the effect of increasing T and the amount of Hâ‚‚S(g) will decrease.
Decreasing the temperature:
The T is a part of the reactants, increasing the T decreases the amount of the reactants. So, the reaction will be shifted to the left to suppress the effect of decreasing T and the amount of Hâ‚‚S(g) will increase.
III. Changing the Hâ‚‚ concentration:
Hâ‚‚ is a part of the products.
Increasing the Hâ‚‚ concentration:
Hâ‚‚ is a part of the products, increasing Hâ‚‚ increases the amount of the products. So, the reaction will be shifted to the left to suppress the effect of increasing Hâ‚‚ and the amount of Hâ‚‚S(g) will increase.
Decreasing the Hâ‚‚ concentration:
Hâ‚‚ is a part of the products, decreasing Hâ‚‚ decreases the amount of the products. So, the reaction will be shifted to the right to suppress the effect of decreasing Hâ‚‚ and the amount of Hâ‚‚S(g) will decrease.
I. Changing the pressure:
Increasing the pressure: the amount of Hâ‚‚S(g) will increase.
Decreasing the pressure: the amount of Hâ‚‚S(g) will decrease.
II. Changing the temperature:
Increasing the temperature: the amount of Hâ‚‚S(g) will decrease.
Decreasing the temperature: the amount of Hâ‚‚S(g) will increase.
III. Changing the Hâ‚‚ concentration:
Increasing the Hâ‚‚ concentration: the amount of Hâ‚‚S(g) will increase.
Decreasing the Hâ‚‚ concentration: the amount of Hâ‚‚S(g) will decrease.
Explanation:
Le Châtelier's principle states that when there is an dynamic equilibrium, and this equilibrium is disturbed by an external factor, the equilibrium will be shifted in the direction that can cancel the effect of the external factor to reattain the equilibrium.
I. Changing the pressure:
When there is an increase in pressure, the equilibrium will shift towards the side with fewer moles of gas of the reaction. And when there is a decrease in pressure, the equilibrium will shift towards the side with more moles of gas of the reaction.
For the reaction: CH₄(g) + 2H₂S(g) ⇄ CS₂(g) + 4H₂(g),
The reactants side (left) has 3.0 moles of gases and the products side (right) has 5.0 moles of gases.
Increasing the pressure: will shift the reaction to the side with lower moles of gas (left side), amount of Hâ‚‚S(g) will increase.
Decreasing the pressure: will shift the reaction to the side with lower moles of gas (right side), amount of Hâ‚‚S(g) will decrease.
II. Changing the temperature
The reaction is endothermic since the sign of ΔH is positive.
So the reaction can be represented as:
CH₄(g) + 2H₂S(g) + heat ⇄ CS₂(g) + 4H₂(g).
Increasing the temperature:
The T is a part of the reactants, increasing the T increases the amount of the reactants. So, the reaction will be shifted to the right to suppress the effect of increasing T and the amount of Hâ‚‚S(g) will decrease.
Decreasing the temperature:
The T is a part of the reactants, increasing the T decreases the amount of the reactants. So, the reaction will be shifted to the left to suppress the effect of decreasing T and the amount of Hâ‚‚S(g) will increase.
III. Changing the Hâ‚‚ concentration:
Hâ‚‚ is a part of the products.
Increasing the Hâ‚‚ concentration:
Hâ‚‚ is a part of the products, increasing Hâ‚‚ increases the amount of the products. So, the reaction will be shifted to the left to suppress the effect of increasing Hâ‚‚ and the amount of Hâ‚‚S(g) will increase.
Decreasing the Hâ‚‚ concentration:
Hâ‚‚ is a part of the products, decreasing Hâ‚‚ decreases the amount of the products. So, the reaction will be shifted to the right to suppress the effect of decreasing Hâ‚‚ and the amount of Hâ‚‚S(g) will decrease.
