question archive The decomposition of is represented by the equation 2HI <---> H2 + I2 The following experiment was devised to determine the equilibrium constant of the reaction
The decomposition of is represented by the equation 2HI <---> H2 + I2 The following experiment was devised to determine the equilibrium constant of the reaction
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The decomposition of is represented by the equation
2HI <---> H2 + I2
The following experiment was devised to determine the equilibrium constant of the reaction.
HI is introduced into five identical 400-cm3 glass bulbs, and the five bulbs are maintained at 623 K.The amount of produced over time is measured by opening each bulb and titrating the contents with 0.0150 M . The reaction of I2 with the titrant is
I2 + 2Na2S2O3 ---> Na2S4O6 +2NaI
Data for the experiment are provided in this table.
Bulb Initial mass of
() Time
() Volume of titrant
()
1 0.300 2 20.96
2 0.320 4 27.90
3 0.315 12 32.31
4 0.406 20 41.50
5 0.280 40 28.68
Part A
In which bulb would you expect the composition of gases to be closest to equilibrium?
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Courtesy of M. Scotty
***sorry no subscripts***
Use the equation
K= [I2][H2]
[HI2]
Step 1. Solve for [I2]
--use M=mol from Na2S2O3---à.0150M=mol à4.30x10^-4 mol
L .02868L
--Based of stoic coefficients
4.30x10^-4 mol x 1molI2 = 2.15x10^-4mol I2
2mol Na2S2O3
--- again M=mol M=2.15x10^-4 à[.0075] for I2
L .02868L
Step 2. [h2]
-- [I2] = [H2] based off stoic coefficients.
Step 3. [HI]
Use the .280g sample and divide by MM to get initial moles
.280g/127.90g = .00219mol (initial moles)
*I cannot remember how I solved for final moles sorry but the valueis .00043 mol
.00219 mol - .00043mol = .00176 mol divide this by vol à .00176mol =.0614
.02868L
.0614 is the [HI]
Step 4. Solve for K
K= [I2][H2] K=[.0075][.0075] à5.63x10^-5 = 1.50x10^-2
[HI2] [.0614]^2 3.77x10^-3
