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zimovet [89]
3 years ago
13

Explain, in terms of particles, concentration, and reaction rate, what you expect to happen when methane gas (CH4) and hydrogen

sulfide gas (H2S) are sealed in a flask and reach a state of equilibrium.
CH4(g) + 2H2S(g) <----> CS2(g) _ 4H2(g)
Chemistry
1 answer:
xeze [42]3 years ago
3 0

Answer:

Constant number of particles and concentration, equal forward and reverse reaction rates

Explanation:

In order to properly answer this question, we need to define the state of equilibrium in terms of its characteristics:

  • at equilibrium, the rate of the forward reaction becomes equal to the rate of the reverse reaction;
  • the molarity of reactants and products remains constant (however, this doesn't imply that the reaction stops knowing the first point of the definition).

Given this equilibrium system, when equilibrium is reached, the moles, the number of particles and the molarity of methane, hydrogen sulfide, carbon disulfide and hydrogen gas, they all become constant and don't change over time. When equilibrium is established, molarity values don't change, and since molarity is directly proportional to the number of moles (as well as the number of atoms), the latter don't change too.

In terms of the reaction rate, it is true for any equilibrium system, including the one defined, that the forward reaction rate (reaction between methane and hydrogen sulfide) becomes equal to the reverse reaction rate (reaction between carbon disulfide and hydrogen gas).

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5 0
3 years ago
Consider an Al-4% Si alloy. Determine (a) if the alloy is hypoeutectic or hypereutectic, (b) the composition of the first solid
ozzi

Answer:

(a) Hypoeutectic

(b) Alpha solid, aluminium

(c) 70% α , 30% β

(d) 97.6% α, 2.4% β

(e) 97.6% α, 2.4% β

(f) 97% α, 3% β

Explanation:

(a) The eutectic composition for Al Si alloy is 11.7 wt% silicon, therefore, an Al-4% Si alloy is hypoeutectic

(b) For the hypoeutectic alloy, aluminium, Al, is expected to form first, such that the aluminium content is reduced till the point it gets to the eutectic proportion of 11.7 wt% silicon

(c) At  578°C we have

% α:  Al      (11 - 4)/(11 - 1) = 70% α

% L:  Si      100 - 70 = 30% β

(d) At  576°C we have

α: 99.83% Si    (99.83 - 4)/(99.83- 1.65) = 97.6% α

β: 1.65% Si (4 - 1.65)/(99.83- 1.65) = 2.4% β

(e) Primary α: 1.65% α (99.83 - 4)/(99.83 - 1.65) = 97.6% α

Eutectic 4% Si  = 100 - 97.6 = 2.4% β

(f) At 25°C we have;

α%: (99.83 - 4)/(99.83 - 1) = 97% α

β%:   100 - 97 = 3% β.

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Explanation:

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