Answer:
we will use the Clausius-Clapeyron equation to estimate the vapour pressures of the boiling ethanol at sea level pressure of 760mmHg:
ln (P2/P1) =
-
)
where
P1 and P2 are the vapour pressures at temperatures T1 and T2
Δ
vapH = the enthalpy of vaporization of the ETHANOL
R = the Universal Gas Constant
In this problem,
P
1
=
100 mmHg
; T
1
=
34.7 °C
=
307.07 K
P
2
=
760mmHg
T
2
=T⁻²=?
Δ
vap
H
=
38.6 kJ/mol
R
=
0.008314 kJ⋅K
-1
mol
-1
ln
(
760/10)=(0.00325 - T⁻²) (38.6kJ⋅mol-1
/0.008314
)
0.0004368=(0.00325 - T⁻²)
T⁻²=0.002813
T² = 355.47K
Explanation:
The reaction equation will be as follows.

Using bond energies, expression for calculating the value of
is as follows.

On reactant side, from
number of bonds are as follows.
C-C bonds = 1
C-H bonds = 6
From
; Cl-Cl bonds = 1
On product side, from
number of bonds are as follows.
C-C bonds = 1
C-H bonds = 5
C-Cl bonds = 1
From HCl; H-Cl bonds = 1
Hence, using the bond energies we will calculate the enthalpy of reaction as follows.

=
= -102 kJ/mol
Thus, we can conclude that change in enthalpy for the given reaction is -102 kJ/mol.
Answer:
A 1.0 g sample of propane, C3H8, was burned in the calorimeter.
The temperature rose from 28.5 0C to 32.0 0C and the heat of combustion 10.5 kJ/g.
Calculate the heat capacity of the calorimeter apparatus in kJ/0C
Explanation:

Given,
The heat of combustion = 10.5kJ/g.

Substitute these values in the above formula to get the value of heat capacity of the calorimeter.

Answer:
The heat capacity of the calorimeter is 