Set the reaction and state conditions
Enter positive stoichiometric numbers. Product and reactant signs are applied automatically.
Default values used by this page
| Input | Default value | Unit or note |
|---|---|---|
| S° NH3(g) | 192.77 | J mol⁻¹ K⁻¹ |
| S° N2(g) | 191.61 | J mol⁻¹ K⁻¹ |
| S° H2(g) | 130.68 | J mol⁻¹ K⁻¹ |
| Cp values | 35.06, 29.12, 28.84 | NH3, N2, H2; J mol⁻¹ K⁻¹ |
| Reaction equation | 1 N2 + 3 H2 → 2 NH3 | Gas phase |
| Reference state | 298.15 K and 1.00 pressure unit | All pressure values must share one unit. |
Formula used
Reference reaction entropy: ΔS°rxn = νNH3S°NH3 − νN2S°N2 − νH2S°H2
Heat-capacity difference: ΔCp = νNH3Cp,NH3 − νN2Cp,N2 − νH2Cp,H2
Temperature adjustment: ΔS°(T) = ΔS°(Tref) + ΔCp ln(T/Tref)
Pressure adjustment: ΔS(T,p) = ΔS°(T) − R ln Q, where Q = (pNH3/p°)νNH3 / [(pN2/p°)νN2(pH2/p°)νH2]
The calculation treats gases as ideal and heat capacities as constant over the selected range. The optional Gibbs energy estimate uses ΔG = ΔH − TΔS.
How to use this calculator
- Confirm the balanced ammonia reaction and keep each coefficient positive.
- Enter standard molar entropy values for ammonia, nitrogen, and hydrogen.
- Enter heat capacities when you need a temperature correction.
- Set the reaction and reference temperatures in kelvin.
- Use one consistent pressure unit for every partial pressure and reference pressure.
- Submit the form. Review the reaction result and the per-mole ammonia result.
- Download a CSV or print the result for your records.
Understanding ammonia formation entropy
Ammonia formation is a useful thermodynamics example. The balanced gas reaction is nitrogen plus hydrogen forming ammonia. Entropy measures how energy and matter are dispersed. It does not simply measure disorder. Each gaseous species has a standard molar entropy. The reaction value combines product and reactant entropy terms using their stoichiometric coefficients.
For the common equation, N2(g) + 3H2(g) → 2NH3(g), four gas moles become two. That reduction often produces a negative reaction entropy. The result is expected because fewer independent gas particles remain after reaction. However, temperature, composition, and pressure can shift the actual entropy change from its standard-state value.
The calculator begins with tabulated standard molar entropy values. It multiplies the ammonia value by its product coefficient. It then subtracts the nitrogen and hydrogen contributions. This gives ΔS° for the balanced reaction at the selected reference temperature. Dividing by the ammonia coefficient gives an entropy change per mole of ammonia formed. This distinction matters when you compare data from different chemical equations.
Heat-capacity values help estimate the temperature effect. Constant-pressure heat capacity describes how much heat a substance absorbs during warming. The reaction heat-capacity difference is calculated using the same coefficients. The calculator adds ΔCp ln(T/Tref) to the reference entropy change. This approximation works best across moderate temperature ranges. Use high-quality temperature-dependent tables for precise engineering or research work.
Pressure inputs apply an ideal-gas correction. Each partial pressure is compared with the chosen standard pressure. The reaction quotient gathers those pressure ratios. The correction is minus R times ln Q. Pure standard-state gases give Q equal to one. Their pressure correction becomes zero. Nonstandard mixtures can therefore change the reaction entropy meaningfully, especially when component pressures differ greatly.
The optional enthalpy input estimates Gibbs energy. The calculator starts with standard formation enthalpy for one mole of gaseous ammonia. It scales that value to the reaction equation. A simple ΔCp(T−Tref) term estimates the enthalpy temperature shift. Gibbs energy is then ΔH minus TΔS. A negative Gibbs energy indicates thermodynamic favorability at the entered state. It does not predict how rapidly ammonia forms.
Use consistent units throughout the calculation. Enter entropies and heat capacities in joules per mole-kelvin. Enter formation enthalpy in kilojoules per mole. Enter temperature in kelvin. Enter every pressure using the same unit as the reference pressure. Keep coefficients positive because the calculator assigns product and reactant signs internally. Verify the balanced equation before interpreting results.
This tool supports teaching, screening, and sensitivity checks. It does not replace validated process simulation. Real ammonia systems may show nonideal behavior at elevated pressure. Fugacity corrections then become more appropriate than simple partial-pressure ratios. Review source data and assumptions before making design, safety, or operating decisions. Save the output table for comparisons across several input conditions. Compare several temperatures and pressures to reveal trends without overlooking the limits of the ideal-gas model used here in practice.
Frequently asked questions
1. Which reaction does the calculator use?
The default reaction is N2(g) + 3H2(g) → 2NH3(g). You may change the positive coefficients when you need another balanced representation of ammonia formation.
2. Why is the entropy change commonly negative?
The default reaction changes four gas moles into two gas moles. Fewer independently moving gas particles often means a lower entropy for the products than for the reactants.
3. Which units should I enter?
Use J/mol·K for entropy and heat capacity. Use kelvin for temperature. Use kJ/mol for ammonia formation enthalpy. Use one shared pressure unit for every pressure field.
4. What does the per-mole ammonia result mean?
It divides the entered reaction entropy by the ammonia coefficient. This gives the entropy change associated with producing one mole of NH3 under the selected conditions.
5. How does temperature affect the calculation?
The calculator estimates a temperature correction from ΔCp ln(T/Tref). This assumes constant heat capacities, so it is most suitable for moderate temperature changes.
6. Why are partial pressures included?
Partial pressures create an ideal-gas correction through the reaction quotient. At the standard pressure for every gas, Q equals one and the pressure correction is zero.
7. Must all pressure fields use the same unit?
Yes. The calculation uses pressure ratios. Bar, atm, and kPa are each acceptable only when the component pressures and reference pressure use the same unit.
8. Can I use this for liquid ammonia?
Not without replacing the gas-phase data and pressure model. This page is designed for gaseous ammonia, nitrogen, and hydrogen under an ideal-gas approximation.
9. What does a negative Gibbs energy estimate show?
It suggests thermodynamic favorability at the entered state. It does not indicate reaction speed, catalyst performance, equipment limits, or whether equilibrium is reached quickly.
10. Is the ideal-gas model always accurate?
No. It is a practical approximation. At high pressure or strong nonideal conditions, fugacity-based methods and validated property data provide more reliable results.
11. Why must coefficients stay positive?
The calculator already treats ammonia as a product and nitrogen and hydrogen as reactants. Positive coefficients prevent accidental sign reversal in the entropy, heat-capacity, and pressure terms.