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Temka [501]
3 years ago
6

What factors affect the dynamic state of equilibrium in a chemical reaction and how?

Chemistry
1 answer:
yanalaym [24]3 years ago
8 0

Answer:

Only changes in temperature will influence the equilibrium constant K_c. The system will shift in response to certain external shocks. At the new equilibrium Q will still be equal to K_c, but the final concentrations will be different.

The question is asking for sources of the shocks that will influence the value of Q. For most reversible reactions:

  • External changes in the relative concentration of the products and reactants.

For some reversible reactions that involve gases:

  • Changes in pressure due to volume changes.

Catalysts do not influence the value of Q. See explanation.

Explanation:

\displaystyle K_c = {e}^{\Delta G/(R\cdot T)}.

Similar to the rate constant, the equilibrium constant K_c depends only on:

  • \Delta G the standard Gibbs energy change of the reaction, and
  • T the absolute temperature (in degrees Kelvins.)

The reversible reaction is in a dynamic equilibrium when the rate of the forward reaction is equal to the rate of the backward reaction. Reactants are constantly converted to products; products are constantly converted back to reactants. However, at equilibrium Q = K_c the two processes balance each other. The concentration of each species will stay the same.

Factors that alter the rate of one reaction more than the other will disrupt the equilibrium. These factors shall change the rate of successful collisions and hence the reaction rate.

  • Changes in concentration influence the number of particles per unit space.
  • Changes in temperature influence both the rate of collision and the percentage of particles with sufficient energy of reaction.

For reactions that involve gases,

  • Changing the volume of the container will change the concentration of gases and change the reaction rate.

However, there are cases where the number of gases particles on the reactant side and the product side are equal. Rates of the forward and backward reaction will change by the same extent. In such cases, there will not be a change in the final concentrations. Similarly, catalysts change the two rates by the same extent and will not change the final concentrations. Adding noble gases will also change the pressure. However, concentrations stay the same and the equilibrium position will not change.

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kvv77 [185]

274 mL H3 O+ and fully neutralized

It will take one teaspoon of Mg(OH)2 to completely neutralize 2.00×10^2mL  of H3O+.

<h3>What is the purpose of milk of magnesia?</h3>
  • For a brief period of time, this medicine is used to relieve sporadic constipation.
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<h3>What dosage of milk of magnesia is recommended for constipation?</h3>
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learn more about milk of magnesia here

brainly.com/question/15178597

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the question you are looking for is

People often take milk of magnesia to reduce the discomfort associated with acid stomach or heartburn. The recommended dose is 1 teaspoon, which contains 4.00x 10^{2} mg of Mg(OH)_2. What volume of an HCl solution with a pH of 1.3 can be neutralized by one dose of milk of magnesia? If the stomach contains 2.00x10^{2}mL of pH 1.3 solution, is all the acid neutralized? If not, what fraction is neutralized?

7 0
1 year ago
How many atoms are in 13.97 liters of water vapor at STP
Arisa [49]
<span>Let's </span>assume that water vapor has ideal gas behavior. <span>
Then we can use ideal gas formula,
PV = nRT<span>

</span><span>Where, P is the pressure of the gas (Pa), V is the volume of the gas (m³), n is the number of moles of gas (mol), R is the universal gas constant ( 8.314 J mol</span></span>⁻¹ K⁻¹) and T is temperature in Kelvin.<span>
<span>
</span>P = 1 atm = 101325 Pa (standard pressure)
V = 13.97 L = 13.97 x 10</span>⁻³ m³<span>
n = ?
R = 8.314 J mol</span>⁻¹ K⁻¹<span>
T = 0 °C = 273 K (standard temperature)
<span>
By substitution,
</span>101325 Pa x 13.97x 10</span>⁻³ m³ = n x 8.314 J mol⁻¹ K⁻¹ x 273 K<span>
                                          n = 0.624 mol
<span>
Hence, the moles of water vapor at STP is 0.624 mol.

According to the </span></span>Avogadro's constant, 1 mole of substance has 6.022 × 10²³ particles.
<span>
Hence, number of atoms in water vapor = 0.624 mol x </span>6.022 × 10²³ mol⁻¹
<span>                                                                = 3.758 x 10</span>²³<span>

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V=(2.10 moles*0.082 (atm l)/º(K mol)*237ºK)  / 1 atm=47.01 litres

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