How would you slove this? The reaction in question will be carried out in a calo
ID: 520765 • Letter: H
Question
How would you slove this?
The reaction in question will be carried out in a calorimeter. The volume of the chamber inside is 2.00 L. Experiment starts by evacuating the chamber to 0.00 kPa. Then oxygen gas is filled into the chamber till pressure 200. kPa. Then hydrogen gas is filled into the same chamber until total pressure 400. kPa. The oxygen and hydrogen mixture is ignited by a spark with 3.00 kJ energy. Calculate the total pressure inside the chamber after reaction. (The heat capacity of this calorimeter is 944.12 J/K. If other constants or equations are needed, you may obtain them from any resources but cite the resource in the discussion board.)
Explanation / Answer
Ans. Step 1. Given, volume of calorimeter chamber = 2.0 L
Pressure due to O2 = 200 kPa = 1.97385 atm ; [1 kPa = 0.00986923 atm]
Equal volume of gas exerts equal pressure at constant T and V.
Pressure due to H2 = 400 kPa – Initial pressure due to O2
= 400 kPa – 200 kPa
= 200 kPa
= 1.97385 atm
So, in the 2.0 L chamber, there is 1.97385 atm each of O2 and H2.
It’s assumed that the initial temperature of the reaction mixture as well as calorimeter chamber is 250C- room temperature.
Now, using Ideal gas equation: PV = nRT - equation 1
Where, P = pressure in atm
V = volume in L
n = number of moles
R = universal gas constant= 0.0821 atm L mol-1K-1
T = absolute temperature (in K) = (0C + 273.15) K
Putting the values in equation 1-
1.97385 atm x 2.0 L = n x (0.0821 atm L mol-1K-1) x 298.15 K
Or, n = (3.9477 atm L) / (24.478115 atm L mol-1) = 0.16127467 mol
Equal moles of two gases exert equal pressure at constant V and T. Therefore, there is 0.16127467 mol of each gas in the chamber.
Step 2. Balanced reaction: 2H2 + O2 ---------> 2H2O
Stoichiometry: 2 moles of H2 react with 1 mol O2 to form 2 mol H2O.
Theoretical stoichiometry ratio of reactants = H2 : O2 = 2 : 1
We have 0.16127467 mol each of the gases.
Experimental stoichiometry ratio of reactants = H2 : O2 = 1 : 1.
Therefore, H2 is the limiting reactant.
Step 3: Molar enthalpy of combustion of H2 = - 286 kJ/ mol. [Ref: http://www.spaceflight.esa.int/impress/text/education/Catalysis/Question_Catalysis_13.html ] Note that the -ve sign indicates the release of heat during reaction.
Total heat produced during combustion =
Molar enthalpy of combustion of H2 x Moles of H2 combusted
(- 286 kJ/ mol) x 0.16127467 mol
= - 46.12455562 kJ
Step 4: Composition of gases after combustion-
All of 0.16127467 mol H2, the liming reactant, is consumed up.
Moles of H2O formed = moles of H2 combusted = 0.16127467 mol
Moles of O2 consumed = (1/2) x moles of H2 combusted = 0.080637335 mol
Moles of O2 remaining = Initial moles of O2 – moles of O2 consumed
= 0.16127467 mol - 0.080637335 mol
= 0.080637335 mol
Total moles remaining after combustion =
Moles of H2O produced + Moles of O2 remaining
= 0.16127467 mol + 0.080637335 mol
= 0.241912005 mol
Step 5: Heat gained by Calorimeter is given by-
Q = C dT ; [where, C = heat capacity of calorimeter]
We have,
Q = 46.12455562 kJ = 46124.55562 J - from #3
dT = T2 – 250C = T2 – 298.15 K
C = 944.12 J/K
Putting the values in equation 2-
46124.55562 J = (944.12 J/K) x (T2 – 298.15 K)
Or, 46124.55562 J / (944.12 J/K) = T2 – 298.15 K
Or, 48.85 K = T2 – 298.15 K
Or, T2 = 48.85 K + 298.15 K = 347.0 K
Therefore, the equilibrium temperature of the chamber after combustion = 347.0 K
Step 6: So far, we have-
Total moles of gases remaining in chamber after combustion = 0.241912005 mol
Temperature of chamber after combustion = 347.0 K
Volume of chamber = 2.0 L
Again, putting the values in equation 1-
P x 2.0 L = 0.241912005 mol x (0.0821 atm L mol-1K-1) x 347.0 K
Or, P = 6.892 atm L / (2.0 L) = 3.446 atm
Therefore, pressure in the chamber after combustion = 3.446 atm
= 349.2 kPa
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