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How much would the concentration of CO2(aq) change due to a temperature change from 25°C to 26°C at 1atm?
I was having a little trouble posting this question, sorry
Any help would be greatly appreciated, thanks
2 Answers
- pisgahchemistLv 77 years agoFavorite Answer
Whoa! Back the boat up. While there seems to be a lot of work there, N3...'s solution makes no sense. He is calculating the moles of GAS in 1 L of CO2 at 298K and 1 atm. That isn't the amount that is dissolved in SOLUTION at 298K.
The amount of CO2 dissolved in water can be determined from Henry's law.
c = kH(p) ..... c is the concentration of the gas, kH is the Henry's law constant for the particular gas and p is the partial pressure of the gas above the solution.
c = 0.034 mol/Latm x 1.00 atm = 0.034 mol/L
The Henry's law constant varies with temperature, and must be calculated at the new temperature.
kH(T) = kH(T') x e^(C(1/T - 1/T')) .... where T is the new temperature, T' is 298K, C is a constant for each gas. For CO2, C = 2400
kH(T) = 0.034 mol/Latm x e^(2400(1/299 - 1/298))
kH(T) = 0.033 mol/Latm
Calculating the concentration at 26C....
c = kH(p)
c = 0.033 mol/Latm x 1.00 atm = 0.033 mol/L
The concentration will decrease by 0.001 mol/L as the CO2 solution goes from 25C to 26C.
- N3XV$81Lv 47 years ago
PV = nRT
n = PV/RT
P = 1atm
R = 0.08205746 Latm/Kmol
T1 = 25C + 237.15K = 298.15K
T2 = 26C + 237.15K = 299.15K
n = PV/RT
n = 1atm(1L)/(0.08205746 Latm/Kmol)(298.15K)
n = 1/24.465mol
n = 0.0409mol CO2 --- initial mass of CO2
n = 1atm(1L)/(0.08205746 Latm/Kmol)(299.15K)
n = 1/24.547mol = 0.0407mol CO2 -- final mass of CO2
dm = mf - mi
where,
M = m/V
M = 0.0409mol/L = 0.0409M <--- Molarity of CO2 at 25C
M = 0.0407mol/L = 0.0407M <---- Molarity of CO2 at 26C
dM = Mf - Mi = 0.0407M - 0.0409M = -0.0002M <---- change of CO2 when T is increased by 1C
therefore,
----- The concentration of CO2 is decreased by about 0.0002M when the temperature is increased from 25C to 26C. The final concentration of CO2 at 26C is 0.0407M.