Figure 2. The "van't Hoff plot" of any substance that has a measurable vapor or sublimation pressure is a straight line in coordinates of absolute pressure logarithm versus inverse absolute temperature. This includes water, dry ice, liquid ammonia and metal hydrides.
The slope of a van't Hoff plot is linearly related to the latent heat (¦¤H) of a vaporizing or subliming substance. The Clausius-Clapeyron equation is:
I. dP/P = ¦¤H dT/RT^{2}
Where R is the universal gas constant, 8.314 Joules per gram-mole of ideal gas per Kelvin.
Integration of Equation I between two points on the van't Hoff plot yields:
ln (P_{2} /P_{1} ) = ln P_{2} - ln P_{1} = -¦¤H/R (1/T_{2} ¨C 1/T_{1} )
Solving for ¦¤H gives:
II. ¦¤H = -R (lnP_{2} ¨C lnP_{1} )/ (1/T_{2} ¨C 1/T_{1} )
From the set of converted data above, let's use the highest pressure and temperature as P_{2} and T_{2} and the lowest pressure and temperature as P_{1} and T_{1} . Substituting these values into equation II gives the latent heat of desorption:
¦¤H des = +31773 Joules per gram-mole of H_{2} .
The plus sign tells us that this much heat must enter the reaction vessel per gram-mole of H_{2} desorbed. The best of metal hydrides are nearly reversible so, for absorption, we may write:
¦¤H abs = -31773 Joules per gram-mole H_{2} .
The sign of the enthalpy change is negative because heat flows out of a reaction vessel during absorption of hydrogen.
Standard Conditions
"Standard" conditions are needed to solve Gibbs' function from a set of experimental pressure-temperature data. The usual standard environment is 25¡æ (298.15K) and 1 atmosphere (760 millimeters of mercury or 101.3 kPa). The conventional notation for standard Enthalpy is ¦¤H^{o}.
We just derived ¦¤H over a range of temperature-not at exactly 298.15K. If we wanted to know ¦¤H^{o} very precisely we would need high precision pressure-temperature data in a narrow range around 298.15K. This is true because ¦¤H is the difference between the enthalpies of the final and initial substances involved in the change. In the case of desorption, the final substances are metal and hydrogen. The initial substance is metal hydride. When we warm or cool a substance its enthalpy changes according to:
¦¤H = m_{p}¦¤T
where m is the mass of substance in grams, c p is the specific heat in Joules per gram per Kelvin at constant pressure and ¦¤T is the temperature difference in Kelvins. This type of enthalpy change is called "sensible heat".
We can ignore sensible heat effects only if we know that the cps of all substances do not change significantly over the range of temperature of our data (298.15K to 338.15K). The specific heat of hydrogen increases from 14.31 to 14.54 over our temperature range-a change of +1.7%. Typical metals vary by twice as much. We do not have c p versus T data for most modern hydride alloys or their hydrides. We cannot believe a 5 significant digit result like ¦¤H^{o} des = 31773 Joules per gram-mole of H_{2} . A detailed error analysis would probably justify a result like:
¦¤H^{o}des = 32¡À1 kJoules per gram-mole of H_{2} .
Standard Gibbs Free Energy Change
The standard Gibbs free energy change is the maximum reversible work that can be obtained from a mole of ideal gas at standard temperature against the resistance of standard atmospheric pressure. Hydrogen is very nearly ideal at moderate pressures and temperatures. Any thermodynamics text will derive
¦¤G^{o} = RT^{o} ln(P/P^{o})
as the standard Gibbs free energy change per mole of ideal gas produced. The metal hydride we tested produces 2.00 atmospheres of hydrogen pressure at 298.15K. Therefore, we calculate as follows:
¦¤G^{o}des = 8.314 Joules/g-mol H_{2}^{.}K x 298.15K x ln(2/1)
= 1718¡À13 Joules/g-mol H_{2} .
The accuracy of this result depends on the accuracy of the pressure and temperature instruments and the ideality of hydrogen gas. If the data are within ¡À0.1K and ¡À0.01 atm, we know ¦¤G^{o}des within ¡À 13 Joules/g-mol H_{2} . ¦¤G^{o}des is positive because the 2.00 atmosphere H_{2} produced has the ability to do work on the 1 atmosphere standard environment.
Standard Entropy
We may now calculate the standard entropy of desorption because it is the only unknown remaining. By rearrangement of Gibbs' function:
¦¤S^{o}des = (¦¤H des^{o} - ¦¤G des^{o})/T^{o},
¦¤S^{o}des = [(32 - 1.7) kJoules/g-mol H_{2} ] / 298.15K = 102¡À4 Joules/g-mol H_{2}^{.}K.
¦¤S^{o}des is positive because a cloud of desorbed gas is more disordered than hydrogen within a solid metal hydride crystal. There was also a small negative change in the entropy of the solid. The net entropy change calculated above is the algebraic sum of the two entropy changes.
Gibbs' Phase Rule
Figure 3 shows the relationship between equilibrium pressure and hydrogen content for an ideal metal hydride at constant temperature. This is called a "pressure versus composition isotherm". |