You are planning to prepare 600 mL of 20% dextrose solution, by mixing your 5% and 50% dextrose solution. How much of each solut
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Answer:
The value of Kc C. remains the same.
The value of Qc C. is less than Kc.
The reaction must: A. run in the forward direction to reestablish equilibrium
The number of moles of Cl2 will B. decrease.
Explanation:
Le Chatelier's Principle states that if a system in equilibrium undergoes a change in conditions, it will move to a new position in order to counteract the effect that disturbed it and recover the state of equilibrium.
A decrease in volume causes the system to evolve in the direction in which there is less volume, that is, where the number of gaseous moles is less.
But temperature is the only variable that, in addition to influencing equilibrium, modifies the value of the constant Kc. So if the volume of the equilibrium system is suddenly decreased at constant temperature: <u><em>The value of Kc remains the same.</em></u>
<u><em> </em></u>As mentioned, if the volume of an equilibrium gas system decreases, the system moves to where there are fewer moles. In this case, being:
PCl₃(g) + Cl₂(g) ⇔ PCl₅(g)
The equilibrium in this case then shifts to the right because there is 1 mole in the term on the right, compared to the two moles on the left. So, <u><em>The reaction must: A. run in the forward direction to reestablish equilibrium</em></u>.
By decreasing the volume, and so that Kc remains constant, being:
where nPCl₅, nPCl₃ and nCl₂ are the moles in equilibrium of PCl₅, PCl₃ and Cl₂
so, the number of moles of Cl₂ should decrease.<u><em>The number of moles of Cl2 will B. decrease.</em></u>
If the reaction quotient is less than the equilibrium constant, Qc <Kc, the system will evolve to the right, the direct reaction prevailing, to increase the concentration of products. So in this case, if the reaction moves to the right, <em><u>the value of Qc C. is less than Kc.</u></em>