Equilibrium Concentrations 1 2 3 4 5 [FeSCN2+] 1.49x10-5 M 4.69x10-5 M 7.47x10-5 M 1.01x10-4 M...

Equilibrium Concentrations	1	2	3	4	5
[FeSCN²⁺]	1.49x10^-5 M	4.69x10^-5 M	7.47x10^-5 M	1.01x10^-4 M	2.65x10^-4 M
[Fe³⁺]	6.0x10^-4 M	6.0x10^-4 M	6.0x10^-4 M	6.0x10^-4 M	6.0x10^-4 M
[SCN^-]	2.0x10^-4 M	4.0x10^-4 M	6.0x10^-4 M	8.0x10^-4 M	1.0x10^-3 M

Equilibrium K_eq

124.2

195.4

207.5

210.4

441.7

Average K_eq: 235.8

Does your data support a complete conversion or partial conversion? Based on your results was the measured K_eq too high or too low? Explain please.

Solutions

Expert Solution

Let me tell you something. based on those results of FeSCN in equilibrium, there's no way that the concentration of Fe is the same for all of them, in equilibrium. Are those the innitial concentrations? if that's so, then the equilibrium concentrations would be:

[Fe] = 6x10^-4 - 1.49x10^-5 = 5.85x10^-4 M

[SCN] = 2x10^-4 - 1.49x10^-5 = 1.85x10^-4 M

Therefore the value of Keq would have to be:

Keq = 1.49x10^-5 / 5.85x10^-4 * 1.85x10^-4 = 137.68

Now, if we apply this to the other reactions, the values of Keq would be:

240.14 ; 270.71 ; 280.98 ; 1076.25

Keq avg = 137.68+270.71+280.98+240.14+1'76.25 / 5 = 402.62

Basing on this result, the conversion is partially complete, because it's an equilibrium reaction. Although the reaction is favored and shift to the products, there's always a remaining of product, so the reaction it's not full complete. We can know this with a too high Keq value, that means that this reaction is forwarded to the products.

Hope this helps

queen_honey_blossom answered 1 year ago

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