Chemistry
posted by Anonymous .
Using the standard reduction potentials, calculate the equilibrium constant for each of the following reactions at 298 K.
10Br + 2MnO4 + 16H+ > 2Mn2+ + 8H20 + 5Br2
Work:
2Br > Br2 + 2e (This would be multiplied by 5 to balance out with the upper equation.)
MnO4 + 8H+ + 5e > Mn2+ +4H20 (This would be multiplied by 2 to balance out with the upper equation.)
The standard reduction potential for Br is +1.065.
The standard reduction potential for MnO4 is +1.51.

Look up that equation, nEF = lnK (I'm not at home and can't confirm that equation so check it out) but that is the one you want to use.

I looked back through my notes and found the following equations:
K = e^(delta G/RT)
Delta G = nFE
I know that E = standard reduction of the reduced  standard reduction of the oxidized
I've tried plugging everything I have in, but I'm not getting the right answer. I can't figure out what it is that I am doing wrong.
I have E = 1.51  1.065 and n = 10 
So I have that
1.51 V  1.065 V = .445 V = E
and that
ΔG =  10 mol * 96485 J / mol*V * .445 V
ΔG = 429358 J
so then plugging in:
K = e^((429358)/(8.314 J/mol*K * 298 K))
K = 1.82 * 10^75
The 'correct' answer appears to be 1.5 * 10^75 , but that's not far off, and my homework program accepted my answer as correct.