Please Help! Thank You
1 answer:
You might be interested in
Answer:
(c) 0.11; (d) -24.5 kJ·mol⁻¹; (e) See below; (f) Kc increases;
(g) No effect on ΔH
Step-by-step explanation:
(c) Kc at 125 °C
In Part (a) one molecule of XY dissociated into one X and one Y.
XY ⇌ X + Y
I: 10 0 0
C: -1 +1 +1
E: 9 1 1
Kc = {[X][Y]}/[XY] = (1 × 1)/9 = ⅑ = 0.11
(d) ΔH
ΔH is a constant that is characteristic of the reaction.
ΔH = -24.5 kJ·mol⁻¹
(e) Effect of temperature on concentrations
If ΔH is negative, the reaction is exothermic.
Heat is a product of the reaction, so we can write the equation as
XY ⇌ X + Y + heat
If we lower the temperature, we are removing heat from the system.
<em>Le Châtelier's Principle</em> states that if you apply a stress to a system at equilibrium, it will respond by trying to relieve the stress.
We applied a stress by removing heat, so the system responds by producing more heat. The position of equilibrium moves to the right, and <em>more products will form</em>.
The diagram might look like the one below.
(f) Effect of temperature on Kc
Kc = [Products]/[Reactants]
We are increasing [Products] and decreasing [Reactants].
If you increase the numerator and decrease the denominator, you i<em>ncrease</em> the value of the quotient.
The value of the equilibrium constant increases when the temperature decreases.
(g) Effect of temperature on ΔH
Decreasing the temperature has no effect on ΔH, because the enthalpies of the reactants and products are properties of the substances themselves. They do not depend on the temperature.
Answer:
Explanation:
Hello,
In this case, we work on a dilution process in which we can state that the moles remain the same after the dilution process. In such a way, we can write:
That in terms of molarities and volumes is:
Whereas is the initial molarity (3.00 M) of the stock solution,
the molarity of the diluted solution (0.289 M),
the aliquot of the stock (concentrated) solution and
the volume of the diluted solution (100 mL), thus, we compute
as required:
Best regards.