Determine the equilibrium constant, Keq, for the reaction


5.3 10–19
18.30
1.7 1054
1.9 1018
5.7 10–55

this is an assignment question- as writen, no values, nothing. any help

User Icon for DrBpb222 DrBpb222 answered
14 years ago

These numbers must refer to paragraphs, problems, chapters, or something of that nature where the full problem is found.

User Icon for lucylu lucylu answered
14 years ago

that is the full question. i cut and pasted it. i have emailed the lectures and have heard nothing...as usual. i have asked them if there should be more infor or a picture or something but nothing! how am i supposed to do it then?

User Icon for lucylu lucylu answered
14 years ago

Dr BOB, i just spoke to my lecturer and he says to use reduction potentials and half equations + Ecall=Ered(cathode)+Ered(anode).

so i get half equ:
2Br->Br2+2e- E=-1.065V
I2(s)+2e- -> 2I- E=0.536V
from this i get Ecell=-1.601V
but i don't know where to go from there?

User Icon for lucylu lucylu answered
14 years ago

i need help. my assignmetn its due in less than 24 hrs. can anyone tell me where i went wrong?

User Icon for Explain Bot Explain Bot answered
11 months ago

To determine the equilibrium constant (Keq) for a reaction, you typically need the balanced chemical equation for the reaction and the corresponding concentrations of the reactants and products at equilibrium. However, in your question, there is no balanced chemical equation or any values provided.

If you have the balanced chemical equation and the concentrations of reactants and products, you can calculate Keq using the formula:
Keq = [C]^c [D]^d / [A]^a [B]^b

Where [A], [B], [C], and [D] represent the concentrations of reactants A, B, C, and D, respectively, and a, b, c, and d represent their respective stoichiometric coefficients from the balanced chemical equation.

Since there is no additional information provided in your question, it is not possible to determine the equilibrium constant, Keq, for the reaction. You may need to refer to your assignment instructions or provide more details for further assistance.