g Consider the following system at equilibrium where H° = -87.9 kJ, and Kc = 83.3 , at 500 K: PCl3 (g) + Cl2 (g) PCl5 (g) If the

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g Consider the following system at equilibrium where H° = -87.9 kJ, and Kc = 83.3 , at 500 K: PCl3 (g) + Cl2 (g) PCl5 (g) If the

TEMPERATURE on the equilibrium system is suddenly decreased : The value of Kc A. Increases B. Decreases C. Remains the same The value of Qc A. Is greater than Kc B. Is equal to Kc C. Is less than Kc The reaction must: A. Run in the forward direction to restablish equilibrium. B. Run in the reverse direction to restablish equilibrium. C. Remain the same. Already at equilibrium. The concentration of Cl2 will: A. Increase. B. Decrease. C. Remain the same. Submit AnswerRetry Entire Group

Chemistry

1 answer:

Ksivusya [100] · Answer 1 · 2022-03-28T20:03:43+03:00

Answer:

C. Remains the same

A. Is greater than Kc

B. Run in the reverse direction to restablish equilibrium.

C. Remain the same

Explanation:

As the reaction is exothermic (ΔH < 0), when occurs, produce heat.

Based on Le Chatelier's principle, if the temperature decreases, <em>the system will produce more product trying to restore the equilibrium</em>

* The value of Kc doesn't change because is the constant of the reaction, Kc. That is,

<h3>C. Remains the same</h3>

* Qc is the ratio between products and reactants. As more product is produced, Qc will increase becomes:

<h3>A. Is greater than Kc</h3>

*After the formation of product, the system must:

<h3>B. Run in the reverse direction to restablish equilibrium. </h3>

Because product is produced

*Cl₂ as reactant is consumed producing more product, decreasing its concentration. But when the equilibrium is restored, its concentration would be the same

<h3>C. Remain the same</h3>