in the iodine clock reaction I need help with understanding how to start the mat

Question

in the iodine clock reaction I need help with understanding how to start the math... It is not that I don't know how I just don't understand how to start...2I- + S2O8^2- --> I2 + 2SO4^2-

I2 + 2S2O3^2- --> 2I- S4O6^2-

equation (7) 1/((a/2)-b) * ln((b(a-2x)/a(b-x)) = kt+C when a doesn't equal 2b

equation (8) 2/(a-2x) = kt + C when a equals 2b

a= initial concentration of KI
b= initial concentration of Na2S2O8 ( I think he meant K2S2O8) or (Na2S2O3...)
x = concentration of I2
(a-2x) and (b-x) = concentration of KI and the concentration of Na2S2O8 (again not sure) at time t

1L .1N KI
1L .1N K2S2O8
2L .01N Na2S2O3


.05N KI and K2S2O8 solutions

include values of t x (a-x) (b-x) k T



first, I did this

x = in L
.009 (5min)
.0216 (10min)
.0197 (17 min)
.0228 (21 min)
.01462 (27 min)
.0262 (33 min)
.00909 (37 min)

then I made in in molarity mol/L by multiplying I2 by the Na2S2O2 concentration of .01 mol/L

.01mol/.009L = 1.11mol/L
.463
.5076
.4386
.68399
.38168
1.1001

then I made the values of (a-x) for .05mol/L KI

.05mol/L - 1.11mol/L = -1.06mol/L
-.413
-.458
-.389
-.634
-.332
-1.05

since the value of x is the same for K2S2O8 of .05mol/L (b-x) has the same values as above

for K

since a=2b

I used equation 8

but here is where I am confused... what do I use for the value of C?

2/(.05-2(1.11)) = K (5min *60 s) + C

What do I do?

Explanation / Answer

at t = 0 x = 0 so value of C= 0 take C=0 and calculate K

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