Antoine equation

The Antoine equation is a mathematical expression (derived from the Clausius-Clapeyron relation) of the relation between the vapor pressure and the temperature of pure substances. The equation was first proposed by Ch. Antoine, a French researcher, in 1888. The basic form of the equation is:


 * $$\log P = A-\frac{B}{C+T}$$

and it can be transformed into this temperature-explicit form:


 * $$T = \frac{B}{A-\log P} - C$$

where: $$P$$ is the absolute vapor pressure of a substance
 * $$T$$ is the temperature of the substance
 * $$A$$, $$B$$ and $$C$$ are substance-specific coefficients (i.e., constants or parameters)
 * $$\log$$ is typically either $$\log_{10}$$ or $$\log_e$$

A simpler form of the equation with only two coefficients is sometimes used:


 * $$\log P = A-\frac{B}{T}$$

which can be transformed to:


 * $$T = \frac{B}{A-\log P}$$

Validity ranges
The Antoine equation cannot be used for the entire vapor pressure range from the triple point to the critical point because it is not flexible enough. Therefore two sets of coefficients are commonly used: one set for vapor pressures at temperatures below the normal boiling point (NBP) and one set for vapor pressures at temperatures above the normal boiling point.

Example sets of coefficients
Table 1 lists Antoine equation coefficients for water and for ethanol with each having two sets of coefficients: one for the temperature range below the normal boiling point (NBP) and one for the temperature range above the NBP. The temperature ranges are denoted by the indicated minimum and maximum temperatures.

The coefficients in Table 1 are for temperatures in °C and absolute pressures in mmHg when using $$\log_{10}$$ as the logarithmic function.

Example calculations
The NBP of ethanol is 78.32 °C. Using the Antoine coefficient range for below the NBP from Table 1, the ethanol vapor pressure at the NBP temperature is:


 * (1a)    $$\log P_\mathrm{10} =  8.20417 - \frac{1642.89}{230.000 + 78.32} = 2.8808$$

and
 * (1b)    $$P = 10^{2.8808} = 760.02\; \mathrm{mmHg}$$

Using the Antoine coefficient range for above the NBP from Table 1, the ethanol vapor pressure at the NBP temperature is:


 * (2a)    $$\log P_\mathrm{10} =  7.68117 - \frac{1332.04}{199.200 + 78.32} = 2.8814$$

and
 * (2b)    $$P = 10^{2.8814} = 760.98\; \mathrm{mmHg}$$

(760 mmHg = 1.000 atm = typical atmospheric pressure at sea level)

This example shows the problem caused by using two different sets of coefficients. The two sets of coefficients give different results at the NBP temperature. This causes problems for computational techniques which rely on a continuous vapor pressure curve.

Two solutions are possible: the first approach uses a single Antoine parameter set over a larger temperature range and accepts the increased deviation between calculated and real vapor pressures. A variant of this single set approach is using a parameter set specially fitted for the desired temperature range. The second approach is to use the equations of Wagner or of the AIChE's Design Institute for Physical Properties (DIPPR).

Units
When using Antoine equation coefficients obtained from the technical literature, care must be taken to ascertain the units involved in such coefficients. Some of the coefficients provided in the technical literature are based on using log10 while others are based on using loge, and some are based on using the Celsius scale for temperatures while others are based on using the Kelvin scale. The pressure basis may be in various units other than mmHg. Occasionally, the coefficient B may be given as negative because the basic form of the Antoine equation (see the introductory paragraph of this article) has been rewritten with the minus sign changed to a plus sign.

It is relatively easy to convert the Antoine coefficients based on using Celsius scale temperatures to make them suitable for using Kelvin scale temperatures. All that is required is to subtract 273.15 from the C coefficient.

It is also relatively easy to convert coefficients based on using log10 and pressures in mmHg to make them suitable for using pressures in Pa. Since an absolute pressure of 101,325 Pa is equivalent to an absolute pressure of 760 mmHg, adding the log10 of (101,325 ÷ 760) to the A coefficient is all that is required:

$$A_\mathrm{Pa} = A_\mathrm{mmHg} + \log_{10}\frac{101325}{760} = A_\mathrm{mmHg} + 2.124903$$

The Antoine coefficients for ethanol (below its NBP) based on using log10, mmHg and °C are:

When the coefficients in Table 2 are converted for using Kelvin scale temperatures and pressures in Pa, they are:

Thus, for an ethanol NBP of 78.32 °C = 351 K and using the converted coefficients in Table 3:


 * (3a)    $$\log_{10}P = 10.32907 - \frac{1642.89}{351.47 - 42.85} = 5.00573$$

and


 * (3b)    $$P = 10^{5.00573} = 101,328\; \mathrm{Pa}$$

To make the coefficients in Table 3 (based on using log10) suitable for using loge, the A and B coefficients must be multiplied by loge(10) = 2.302585 to obtain:

Now calculating the vapor pressure using the loge yields:


 * (4a)    $$\log_e P = 23.7836 - \frac{3782.89}{351.47 - 42.85} = 11.52616$$

and


 * (4b)    $$P = e^{11.52616} = 101,332\; \mathrm{Pa}$$

(The 0.006% difference in the calculated vapor pressures between using the log10 coefficients and the loge coefficients is due to the rounding off of numbers used in the calculations.)

Sources of Antoine equation coefficients

 * The NIST online chemistry web book.


 * The online Physical Properties Sources Index (PPSI) of the Swiss Federal Institute of Technology.


 * The online Dortmund Data Bank.


 * Several books and publications.