Chemistry, 30.01.2020 23:59 nolz4066

B. hydrogen sulfide, h2s, is a contaminant in natural gas. it can be removed with the reaction ch4(g) + 2h2s(g) cs2(g) + 4h2(g), δh = 232 kj/mol. how should the following conditions of this reaction be changed to optimize the removal of h2s? explain your answers.

i. changing the pressure

ii. changing the temperature

iii. changing the h2 concentration

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Answer from: ayoismeisalex

I. Changing the pressure:

Increasing the pressure: the amount of H₂S(g) will increase.Decreasing the pressure: the amount of H₂S(g) will decrease.

II. Changing the temperature:

Increasing the temperature: the amount of H₂S(g) will decrease.Decreasing the temperature: the amount of H₂S(g) will increase.

III. Changing the H₂ concentration:

Increasing the H₂ concentration: the amount of H₂S(g) will increase.Decreasing the H₂ concentration: the amount of H₂S(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 H₂S(g) will increase.Decreasing the pressure: will shift the reaction to the side with lower moles of gas (right side), amount of H₂S(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 H₂S(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 H₂S(g) will increase.

III. Changing the H₂ concentration:

H₂ is a part of the products.

Increasing the H₂ concentration:

H₂ is a part of the products, increasing H₂ increases the amount of the products. So, the reaction will be shifted to the left to suppress the effect of increasing H₂ and the amount of H₂S(g) will increase.

Decreasing the H₂ concentration:

H₂ is a part of the products, decreasing H₂ decreases the amount of the products. So, the reaction will be shifted to the right to suppress the effect of decreasing H₂ and the amount of H₂S(g) will decrease.

Answer from: Quest

because u need two or more forms of the same element that contain equal numbers of protons but different numbers of neutrons in their nuclei to form isotopes.

so since they all have the same shape? i mean maybe because m& m and reeses are both chocolates

Answer from: Quest

if you're asking what comes out, it's steam.