October 24, 2019April 6, 2022 by gary_mail@hotmail.com

Standard conditions

The value of the change in enthalpy of a reaction, ΔH_r, depends on the following physical properties of reactants and products:

1. Temperature
2. Pressure or concentration
3. State of matter

In other words, ΔH_rfor a reaction carried out, for example at 298.15 K, is different from that at 373.15K.

To avoid confusion, thermodynamic calculations are often made using data derived under a specific set of conditions known as standard conditions, which are defined as:

1. Temperature: 298.15K (data are sometimes derived at other temperatures)
2. Pressure: 100 kPa or 1 bar
3. Concentration: 1 M
4. State of matter: Each substance is in its normal state (s, l or g) at 100 kPa and 298.15K

For example, the normal state of molecular oxygen at 100 kPa and 298.15K is O₂ (g). Thermodynamic properties, e.g. ΔH_r, that are calculated using standard conditions data are given an additional symbol “o” as a superscript to their current symbols, i.e. ΔH_r^o. ΔH_r^ois therefore the standard enthalpy change of reaction. If ΔH_r^o is observed at a temperature other than 298.15K, the symbol ΔH_r^o(T) will be used, where T is the temperature of the reaction at which ΔH_r^ois observed.

Question

The standard enthalpy change of the manufacture of ammonia via the Haber process, $\frac{1}{2}N_2(g)+\frac{3}{2}H_2(g)\rightleftharpoons NH_3(g)$ , is ΔH_r^o = -46.1 kJmol^-1. However, the process is too slow at 298.15 K and 1 bar. Instead, it takes place at around 720 K and 200 bar. How then is ΔH_r^oobtained?

Answer

The enthalpy change of the reaction, ΔH_r, is first measured at the optimum conditions of 720 K and 200 bar and subsequently converted to ΔH_r^o at 298.15 K and 1 bar using Kirchhoff’s law (the understanding of this law requires the knowledge of chemical thermodynamics at the advanced level).

The value of the standard enthalpy change of a reaction is quoted to match the stoichiometric coefficients of the written reaction equation. For example, if the equation for the Haber process is written as $2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)$ , the quoted standard enthalpy change of reaction is:

$2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)\; \; \; \; \; \; \; \; \Delta H=+92.2\: kJmol^{-1}$

Question

Why is water always in the liquid state in combustion equations like $CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(l)$ when it should be in vapour form considering the temperature of combustion?

Answer

Equations like the combustion of methane are usually written under standard conditions, i.e. with all reactants and products in their normal states at 100 kPa and 298.15K. Although it is not wrong to write the equation as $CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)$ , calculations are often carried out using thermodynamic properties that are quoted in standard conditions, which makes $CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(l)$ more relevant.

Next article: Types of enthalpy changes

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