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CaHeK987 [17]
3 years ago
10

Discussion Questions 1. Given the equilibrium equation for a general reaction, A + B  C + D, explain what happens to the reacta

nts and products from time 0 until the time at which the reaction reaches equilibrium. 2. Given this reaction at equilibrium, C(s) + CO2(g)  2CO(g) (∆H° = 119kJ), explain the changes that would occur when the following stresses are applied or removed for this reaction. (a) CO is removed. (b) Heat is added. (c) CO2 is added. (d) Heat is removed.
Chemistry
1 answer:
ANTONII [103]3 years ago
8 0

Answer:The equilibrium constant for a given reaction is [concentration of products]/[concentration of reactants].

Explanation:

Equilibrium constant=[concentration of products]/[concentration of reactants]

The concentration of reactant  molecules is maximum at time 0 and it  decreases as the reaction proceeds, The concentration of product molecules increases.At equilibrium the concentration of reactants and products are equal.

All the changes would occur in accordance with the LeChateliers principle.

For the given reaction the following changes would occur:

a When CO is removed from the reaction mixture so the reaction would shift towards right that is in forward direction as we are decreasing the concentration of CO so the system would try to increase the concentration of CO and that can happen by more production of CO.

b Since the above reaction is an endothermic reaction so when we would be adding heat to the system that is when we would increase the temperature  the reaction would shift forwards as  more heat energy is absorbed by reactants  to form more products.

c When more CO₂ is added so more amount of reactants are added to the system so the system would try to decrease the amount of reactants that is CO₂ and hence more amount of products would be formed.The reaction would shift in forward direction.

d Since  this reaction is endothermic in nature so when we remove the heat from reaction hence even less amount of heat is present in the system and so the reaction shift in backward direction as the reaction cannot proceed without enough amount of heat.

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The explanation for the large space between particles in a gas is provided by the kinetic theory (option C).

<h3>What is the kinetic theory?</h3>

The kinetic theory is a scientific theory used in physics to understand the behavior of gases.

This theory (kinetic theory) indicates that molecular distances in gases are small when compared to the molecular sizes.

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1 year ago
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Option E, Real gas particles have more complex interactions than ideal gas particles.

In ideal gases, there is absolutely no interaction between any atoms. At all. Atoms simply don't bump into each other in ideal gases.

Obviously, you know that's unrealistic. In real gases, atoms collide into each other all the time.

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lawyer [7]
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A 10.0 mL sample of 0.25 M NaOH(aq) is titrated with 0.10 M HCl(aq) (adding HCl to NaOH). Determine which region on the titratio
Anna11 [10]

Answer:

1) After adding 15.0 mL of the HCl solution, the mixture is before the equivalence point on the titration curve.

2) The pH of the solution after adding HCl is 12.6

Explanation:

10.0 mL of 0.25 M NaOH(aq) react with 15.0 mL of 0.10 M HCl(aq). Let's calculate the moles of each reactant.

nNaOH=\frac{0.25mol}{L} .10.0 \times 10^{-3} L=2.5 \times 10^{-3}mol

nHCl=\frac{0.10mol}{L} \times 15.0 \times 10^{-3} L=1.5 \times 10^{-3}mol

There is an excess of NaOH so the mixture is before the equivalence point. When HCl completely reacts, we can calculate the moles in excess of NaOH.

                    NaOH       +       HCl       ⇒       NaCl      +         H₂O

Initial          2.5 × 10⁻³         1.5 × 10⁻³               0                      0

Reaction    -1.5 × 10⁻³        -1.5 × 10⁻³          1.5 × 10⁻³          1.5 × 10⁻³

Final            1.0 × 10⁻³               0                 1.5 × 10⁻³          1.5 × 10⁻³

The concentration of NaOH is:

[NaOH]=\frac{1.0 \times 10^{-3} mol }{25.0 \times 10^{-3} L} =0.040M

NaOH is a strong base so [OH⁻] = [NaOH].

Finally, we can calculate pOH and pH.

pOH = -log [OH⁻] = -log 0.040 = 1.4

pH = 14 - pOH = 14 - 1.4 = 12.6

5 0
3 years ago
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