https://padlet.com/saunders_r/alkzrxvinxy7 en-us 2020-03-18 06:50:29 UTC 2023-06-25 14:21:07 UTC hello@padlet.com 1238781 https://padlet.com/saunders_r/alkzrxvinxy7/wish/465738782 Kstab:

[CuCl42-]/[cl-]^4 X [Cu(H2O)62+]

Lies  to the rhs

Change this value by changing concentration of reactants/ products


Change temperature  or pressure

]]> 2020-03-19 04:39:33 UTC https://padlet.com/saunders_r/alkzrxvinxy7/wish/465738782 https://padlet.com/saunders_r/alkzrxvinxy7/wish/465738812 2020-03-19 04:39:39 UTC https://padlet.com/saunders_r/alkzrxvinxy7/wish/465738812 https://padlet.com/saunders_r/alkzrxvinxy7/wish/465739000 Kstab = [CuCl42-]/[Cu(H2O)6 2+] [Cl-]^4

FOr 4x10^5, it will lie on the the right hand side

To change the value by changing the concentration of the reactants/products. Change the temperarture.]]> 2020-03-19 04:40:19 UTC https://padlet.com/saunders_r/alkzrxvinxy7/wish/465739000 https://padlet.com/saunders_r/alkzrxvinxy7/wish/465739010 1. Kstab: [CuCl4]2- / [Cu(H2O)62+][Cl-]4

2. Large Ksp value so equilibrium lies to the right 

3. Change the concentration and temperature]]> 2020-03-19 04:40:22 UTC https://padlet.com/saunders_r/alkzrxvinxy7/wish/465739010 https://padlet.com/saunders_r/alkzrxvinxy7/wish/465739057 Kstab = [CuCl4 2-] / [Cu(H2O)6 2+] [Cl- 4]

If it is 4x10^5, it means it lies to the RHS as it is a positive value. There is more CuCl4 than Cu(H2O)6

To change the value, you shift the position of the equilibrium which you can achieve by changing the concentration of CuCl4 or Cu(H2O)6. Changing the temperature can also shift it like how temperature affects Kc

]]> 2020-03-19 04:40:35 UTC https://padlet.com/saunders_r/alkzrxvinxy7/wish/465739057 https://padlet.com/saunders_r/alkzrxvinxy7/wish/465739063 Kstab = [CuCl4 ²⁻] / [Cu(H2O)6 ²⁺][Cl-]⁴

Since 4x10⁵ is greater than 1, it means that the concentration of [CuCl4 ]]]> 2020-03-19 04:40:36 UTC https://padlet.com/saunders_r/alkzrxvinxy7/wish/465739063 https://padlet.com/saunders_r/alkzrxvinxy7/wish/465739086
Kstab= 4x10^5 means equilibrium lies to the right since it is greater than 1, ie at equilibrium the  (multiplication) product of product concentrations is 4x10^5 greater than that of the reactants. Thus [Cu(H2O)6 2+] more stable than [CuCl42-] 

Change value of Kstab by increasing temp (if forward reaction exothermic, this causes Kstab to decrease) and vv.

Change position of eq by increasing concentration of Cl- (shifts left) and [CuCl4 2-] which shifts eq to RHS ]]> 2020-03-19 04:40:40 UTC https://padlet.com/saunders_r/alkzrxvinxy7/wish/465739086 https://padlet.com/saunders_r/alkzrxvinxy7/wish/465739089 Kstab = [[CuCl.4]^2- (aq)]/ [[Cu(H2O)6 2+] [Cl-(aq)]^4 
Kstab > 1 therefore equilibrium lies on the RHS
Change temperature
Change conc]]> 2020-03-19 04:40:41 UTC https://padlet.com/saunders_r/alkzrxvinxy7/wish/465739089 https://padlet.com/saunders_r/alkzrxvinxy7/wish/465739095 1.Kstab = [CuCl4 2-]/ [Cu(H2O)6 2+][Cl-]^4

2.Since 4x10^5 is large, reaction lies to the right. 

3.change temperature

4. change concentration of water


]]> 2020-03-19 04:40:42 UTC https://padlet.com/saunders_r/alkzrxvinxy7/wish/465739095 https://padlet.com/saunders_r/alkzrxvinxy7/wish/465739198 Kstab = [CuCl4 2- (aq)] / [Cu(H2O)6 2+ (aq)] [Cl- (aq)]^4

Kstab is large, so the equilibrium lies to right.

Change the temperature.

Change concentration of reactants. Increase: equilibrium shifts right. Decrease: equilibrium shifts left.]]> 2020-03-19 04:41:05 UTC https://padlet.com/saunders_r/alkzrxvinxy7/wish/465739198 https://padlet.com/saunders_r/alkzrxvinxy7/wish/465739296 Kstab = [CuCl4 2-] / [ Cu(H2O)6 2+] [ Cl-]^4
Large Kstab means equilibrium lies on the right hand side
Kstab can be changed if temperature changes
Position of equilibrium can be changed if the concentration of reactants/products are changed ]]> 2020-03-19 04:41:27 UTC https://padlet.com/saunders_r/alkzrxvinxy7/wish/465739296 https://padlet.com/saunders_r/alkzrxvinxy7/wish/465739538 Kstab = [CuCl42-]/[Cl]^4 [Cu(H2O)6 2+]

The equilibrium lies to the right hand side.
Increase concentration of CuCl4 so equilibrium shifts left/decrease concentration of CL- or the hexaqua chloride.

]]> 2020-03-19 04:42:12 UTC https://padlet.com/saunders_r/alkzrxvinxy7/wish/465739538 https://padlet.com/saunders_r/alkzrxvinxy7/wish/465739603 [CuCl42-] / [Cu(H20)62+] x [Cl-]4 

Since Kstab is positive, it means that eqm lies to the RHS 

By changing the concentration of water, temperature  you can shift equilibrium ]]> 2020-03-19 04:42:25 UTC https://padlet.com/saunders_r/alkzrxvinxy7/wish/465739603 https://padlet.com/saunders_r/alkzrxvinxy7/wish/465739767 Kstab = [CuCl4 2-]/ [Cu(H2O)6 2+] [Cl-]4 
Equilibrium lies on the right hand side of the equation.
Change the conc of reactants and products
Add water ]]> 2020-03-19 04:43:02 UTC https://padlet.com/saunders_r/alkzrxvinxy7/wish/465739767 https://padlet.com/saunders_r/alkzrxvinxy7/wish/465739925 Kstab = [CuCl4 2-] / [Cu(H2O)6 2+] [Cl-]^4
K stab is large, therefore equilibrium lies on the RHS
changing temperature of solution
changing concentrations of reactants or products]]> 2020-03-19 04:43:37 UTC https://padlet.com/saunders_r/alkzrxvinxy7/wish/465739925 https://padlet.com/saunders_r/alkzrxvinxy7/wish/465739938 Kstab = [CuCl4 2-] / [Cu(H2O)6 2+][Cl -]^4

Kstab >1, hence eq lies on the RHS

The olive-green soln turns pale blue (depending on the conc, "olive-green" could be yellow-brown.)


]]> 2020-03-19 04:43:40 UTC https://padlet.com/saunders_r/alkzrxvinxy7/wish/465739938 https://padlet.com/saunders_r/alkzrxvinxy7/wish/465740793 (1)
Kstab=[CuCl4,2-]/[Cu(H20)6,2+][Cl-]4
(2)
Equilibrium lies to RHS due to kstab>1
(3)
Altering concentration of one of the reactants
Changing temperature
(4)
Same as 3]]> 2020-03-19 04:46:30 UTC https://padlet.com/saunders_r/alkzrxvinxy7/wish/465740793