Answer:
a) 48KJ
b) -48KJ
Explanation:
Given that;
ln(K2/K1) = ΔH°/R(1/T2 - 1/T1)
K2= equilibrium constant at T2
K1 = equilibrum constant at T1
R = gas constant
T1 = initial temperature
T2 = final temperature
When we double the equilibrium constant K1; K2 = 2K1
T1 = 310 K
T2 = 310 + 15 = 325 K
ln(2K1/K1) =- ΔH°/R(1/T2 - 1/T1)
ln2 = -ΔH°/8.314(1/325 - 1/310)
0.693 = -ΔH°/8.314(3.08 * 10^-3 - 3.2 * 10^-3)
0.693 = -ΔH°/8.314 (-0.00012)
0.693 = 0.00012ΔH°/8.314
0.693 * 8.314 = 0.00012ΔH°
ΔH° = 0.693 * 8.314/0.00012
ΔH° = 48KJ
b) K2 =K1/2
ln(K1/2/K1) =- ΔH°/R(1/T2 - 1/T1)
ln (1/2) = -ΔH°/8.314 (1/325 - 1/310)
-0.693 = -ΔH°/8.314 (-0.00012)
-0.693 = 0.00012ΔH°/8.314
-0.693 * 8.314 = 0.00012ΔH°
ΔH°= -0.693 * 8.314/0.00012
ΔH°= -48KJ
Answer:
Mole Fraction (H₂O) = 0.6303
Mole Fraction (C₂H₅OH) = 0.3697
Explanation:
(Step 1)
Calculate the mole value of each substance using their molar masses.
Molar Mass (H₂O): 2(1.008 g/mol) + 15.998 g/mol
Molar Mass (H₂O): 18.014 g/mol
200.0 g H₂O 1 mole
--------------------- x ------------------ = 11.10 moles H₂O
18.014 g
Molar Mass (C₂H₅OH): 2(12.011 g/mol) + 6(1.008 g/mol) + 15.998 g/mol
Molar Mass (C₂H₅OH): 46.068 g/mol
300.0 g C₂H₅OH 1 mole
---------------------------- x -------------------- = 6.512 moles C₂H₅OH
46.068 g
(Step 2)
Using the mole fraction ratio, calculate the mole fraction of each substance.
moles solute
Mole Fraction = ------------------------------------------------
moles solute + moles solvent
11.10 moles H₂O
Mole Fraction = -------------------------------------------------------------
11.10 moles H₂O + 6.512 moles C₂H₅OH
Mole Fraction (H₂O) = 0.6303
6.512 moles C₂H₅OH
Mole Fraction = -------------------------------------------------------------
11.10 moles H₂O + 6.512 moles C₂H₅OH
Mole Fraction (C₂H₅OH) = 0.3697
