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Carbon disulfide has a vapor pressure of 363 torr at 25 °C and a normal boiling point of 46.3 °C. Find ΔHvapΔHvap for carbon disulfide.

Question

Carbon disulfide has a vapor pressure of 363 torr at 25 °C and a normal boiling point of 46.3 °C. Find ΔHvapΔHvap for carbon disulfide.

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Step 1

Clausius-Clapeyron equation signifies the relationship between heat of vaporisation and pressure at a specific temperature, labelled below as equation (1). In this equation, P1 represents vapor pressure at temperature T1, P2 represents vapor pressure at temperature T2, R represents universal gas constant and ΔHvap represents enthalpy change of vaporisation of the substance.

Normal boiling point of a liquid is defined as a temperature at which the vapor pressure of liquid becomes equal to the pressure of gas above it (atmospheric pressure). Hence, the liquid has a pressure of 1 atm ( = 760 torr) at its normal boiling point.

ΔΗ
vap
Р.
In
1
1
(1)
Т, т,
R
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ΔΗ vap Р. In 1 1 (1) Т, т, R

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Step 2

Convert the values of temperatures T1 and T2 into Kelvin units by using the above relation.

T, (K)(25+273) K
298 K
T2 (K)(46.3 273) K
-319.3 K
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T, (K)(25+273) K 298 K T2 (K)(46.3 273) K -319.3 K

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Step 3

The enthalpy of vaporisation of carbon disulf...

ΔΗ.
vap
1
1
760 torr
In
363 torr
319.3 K
8.314x10 kJ/molK
298 K
ΔΗ.
vap
(0.000224)
0.739=
8.314x10 kJ/mol
0.739 x 8.314 x103 kJ/mol
ΔΗ,
vap
0.000224
= 27.43 kJ/mol
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ΔΗ. vap 1 1 760 torr In 363 torr 319.3 K 8.314x10 kJ/molK 298 K ΔΗ. vap (0.000224) 0.739= 8.314x10 kJ/mol 0.739 x 8.314 x103 kJ/mol ΔΗ, vap 0.000224 = 27.43 kJ/mol

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