Hydrogen iodide gas decomposes to hydrogen gas and iodine gas: 2 HI ( g ) ⇌ H 2 ( g ) + I 2 ( g ) To determine the equilibrium constant of the system, identical one-liter glass bulbs are filled with 3.20 g of HI and maintained at a certain temperature. Each bulb is periodically opened and analyzed for iodine formation by titration with sodium thiosulfate, Na 2 S 2 O 3 . I 2 ( a q ) + 2 S 2 O 3 2 − ( a q ) → S 4 O 6 2 − ( a q ) + 2 I − ( a q ) It is determined that when equilibrium is reached, 37.0 mL of 0.200 M Na 2 S 2 O 3 is required to titrate the iodine. What is K at the temperature of the experiment?
Hydrogen iodide gas decomposes to hydrogen gas and iodine gas: 2 HI ( g ) ⇌ H 2 ( g ) + I 2 ( g ) To determine the equilibrium constant of the system, identical one-liter glass bulbs are filled with 3.20 g of HI and maintained at a certain temperature. Each bulb is periodically opened and analyzed for iodine formation by titration with sodium thiosulfate, Na 2 S 2 O 3 . I 2 ( a q ) + 2 S 2 O 3 2 − ( a q ) → S 4 O 6 2 − ( a q ) + 2 I − ( a q ) It is determined that when equilibrium is reached, 37.0 mL of 0.200 M Na 2 S 2 O 3 is required to titrate the iodine. What is K at the temperature of the experiment?
Solution Summary: The author explains that the equilibrium constant for the reaction needs to be determined. The number of moles of gas can be calculated from pressure, V, R, Universal gas constant, and T.
Hydrogen iodide gas decomposes to hydrogen gas and iodine gas:
2
HI
(
g
)
⇌
H
2
(
g
)
+
I
2
(
g
)
To determine the equilibrium constant of the system, identical one-liter glass bulbs are filled with 3.20 g of HI and maintained at a certain temperature. Each bulb is periodically opened and analyzed for iodine formation by titration with sodium thiosulfate, Na2S2O3.
I
2
(
a
q
)
+
2
S
2
O
3
2
−
(
a
q
)
→
S
4
O
6
2
−
(
a
q
)
+
2
I
−
(
a
q
)
It is determined that when equilibrium is reached, 37.0 mL of 0.200 M Na2S2O3 is required to titrate the iodine. What is K at the temperature of the experiment?
Calculate the solubility at 25 °C of AgBr in pure water and in 0.34 M NaCN. You'll probably find some useful data in the ALEKS Data resource.
Round your answer to 2 significant digits.
Solubility in pure water:
Solubility in 0.34 M NaCN:
7.31 × 10
M
x10
Ом
Differentiate between normal spinels and inverse spinels.
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Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell