Tabulated below are initial rate data for the reaction ...

[bobypop]

Question 1

Tabulated below are initial rate data for the reaction:
              2Fe(CN)₆^3– + 2I^–→ 2Fe(CN)₆^4– + I₂

Run	[Fe(CN)63–]0	[I–]0	[Fe(CN)64–]0	[I2]0	Rate (M/s)
1	0.01	0.01	0.01	0.01	1 × 10–5
2	0.01	0.02	0.01	0.01	2 × 10–5
3	0.02	0.02	0.01	0.01	8 × 10–5
4	0.02	0.02	0.02	0.01	8 × 10–5
5	0.02	0.02	0.02	0.02	8 × 10–5

What is the value of k?

◦ 10⁷M^–5 s^–1
◦ 10³M^–3 s^–1
◦ 10 M^–2 s^–1
◦ 50 M^–2 s^–1
◦ None of these are correct.

Question 2

Tabulated below are initial rate data for the reaction
              2Fe(CN)₆^3– + 2I^–→ 2Fe(CN)₆^4– + I₂

					Initial
Run	[Fe(CN)63–]0	[I–]0	[Fe(CN)64–]0	[I2]0	Rate (M/s)
1	0.01	0.01	0.01	0.01	1 × 10–5
2	0.01	0.02	0.01	0.01	2 × 10–5
3	0.02	0.02	0.01	0.01	8 × 10–5
4	0.02	0.02	0.02	0.01	8 × 10–5
5	0.02	0.02	0.02	0.02	8 × 10–5

The experimental rate law is:

◦ = k[Fe(CN)₆^3–]²[I^–]²[Fe(CN)₆^4–]²[I₂]
◦ = k[Fe(CN)₆^3–]²[I^–][Fe(CN)₆^4–][I₂]
◦ = k[Fe(CN)₆^3–)]²[I^–]
◦ = k[Fe(CN)₆^3–][I^–]²

◦ = k[Fe(CN)₆^3–][I^–] [Fe(CN)₆^4–]

[princessflame2016]

A FREE account is required to view all solutions! Click to Register

Login with Facebook Login with Google Login with Email

Lorsum iprem. Lorsus sur ipci. Lorsem sur iprem. Lorsum sur ipdi, lorsem sur ipci. Lorsum sur iprium, valum sur ipci et, vala sur ipci. Lorsem sur ipci, lorsa sur iprem. Valus sur ipdi. Lorsus sur iprium nunc, valem sur iprium. Valem sur ipdi. Lorsa sur iprium. Lorsum sur iprium. Valem sur ipdi. Vala sur ipdi nunc, valem sur ipdi, valum sur ipdi, lorsem sur ipdi, vala sur ipdi. Valem sur iprem nunc, lorsa sur iprium. Valum sur ipdi et, lorsus sur ipci. Valem sur iprem. Valem sur ipci. Lorsa sur iprium. Lorsem sur ipci, valus sur iprem. Lorsem sur iprem nunc, valus sur iprium.

Answer Preview

Only 29% of students answer this correctly

[bobypop]

Excellent

[nguyenhoanhat]

Thanks for the timely response, appreciate it