A chemical reaction is the process by which a CHANGE takes place (?)
Answer:
the heat of formation of isopropyl alcohol is -317.82 kJ/mol
Explanation:
The heat of combustion of isopropyl alcohol is given as follows;
C₃H₇OH (l) +(9/2)O₂ → 3CO₂(g) + 4H₂O (g)
The heat of combustion of CO₂ and H₂O are given as follows
C (s) + O₂ (g) → CO₂(g) = −393.50 kJ
H₂ (g) + 1/2·O₂(g) → H₂O (l) = −285.83 kJ
Therefore we have
3CO₂(g) + 4H₂O (g) → C₃H₇OH (l) +(9/2)O₂ which we can write as
3C (s) + 3O₂ (g) → 3CO₂(g) = −393.50 kJ × 3 =
4H₂ (g) + 2·O₂(g) → 4H₂O (l) = −285.83 kJ × 4
3CO₂(g) + 4H₂O (g) → C₃H₇OH (l) +(9/2)O₂ = +2006 kJ/mol
-1180.5 - 1143.32 +2006 = -317.82 kJ/mol
Therefore, the heat of formation of isopropyl alcohol = -317.82 kJ/mol.
Answer:
The partial pressure of Ar is 356.04 mm Hg (= 0.4685 atm)
Explanation:
<u>Step 1:</u> Data given
A mixture of three gases has a total pressure of 1380 mm Hg (=1.81579 atm) at 298 K
Moles of CO2 = 1.27 moles
Moles of CO = 3.04 moles
Moles of Ar = 1.50 moles
<u>Step 2:</u> Calculate total number of moles
Total number of moles = n(CO2)+ n(CO)+ n(Ar) = 1.27 mol+ 3.04 mol+ 1.50 mol = 5.81 moles
<u>Step 3:</u> Calculate mol fraction Ar
Mol fraction Ar = 1.50 mol/5.81 mol = 0.258
<u>Step 4</u>: Calculate partial pressure
1380 mm Hg * 0.258 moles Ar = 356.04 mm Hg = 0.4685 atm
The partial pressure of Ar is 356.04 mm Hg (= 0.4685 atm)
