Calculate Delta G Reaction Using the Following Information | Gibbs Free Energy Calculator


Calculate Delta G Reaction Using the Following Information

Expert Tool for Thermodynamic Spontaneity Analysis


Enter the heat released (negative) or absorbed (positive).
Please enter a valid enthalpy value.


Enter the change in molecular disorder. Note: typically in J, not kJ.
Please enter a valid entropy value.



Temperature at which the reaction occurs.
Temperature must be above absolute zero.


Gibbs Free Energy Change (ΔG)

-5.28 kJ/mol
Reaction is Spontaneous

Temperature in Kelvin (T)

298.15 K

T × ΔS Term

-44.72 kJ/mol

Thermodynamic State

Exergonic

Formula used: ΔG = ΔH – (T × ΔS / 1000)

ΔG vs. Temperature Trend

This chart illustrates how spontaneity changes as temperature increases for your specific ΔH and ΔS values.

Condition Value Description
Standard State 298.15 K Calculation at 25°C
ΔS in kJ/mol·K -0.150 Entropy converted to kJ for ΔG compatibility
T × ΔS Contribution -44.72 kJ/mol Energy associated with disorder change

What is calculate delta g reaction using the following information?

To calculate delta g reaction using the following information is to determine the change in Gibbs Free Energy, which indicates whether a chemical process will occur spontaneously under constant pressure and temperature. In thermodynamics, Gibbs Free Energy (ΔG) represents the “useful” energy available to do work. Researchers, chemists, and students use this value to predict the direction of chemical reactions without needing to measure the entropy of the entire universe.

Common misconceptions include the idea that a negative ΔG means a reaction happens quickly. In reality, ΔG only tells us about the feasibility (thermodynamics), not the speed (kinetics). A reaction could have a very negative ΔG but occur so slowly that it is effectively non-existent without a catalyst.

calculate delta g reaction using the following information Formula and Mathematical Explanation

The primary equation used to calculate delta g reaction using the following information is the Gibbs-Helmholtz equation:

ΔG = ΔH – TΔS

To use this formula correctly, follow these steps:

  • Identify ΔH: The enthalpy change, usually in kilojoules per mole (kJ/mol).
  • Identify ΔS: The entropy change, usually in joules per mole-Kelvin (J/mol·K).
  • Convert ΔS to kJ: Divide ΔS by 1,000 to match the units of ΔH.
  • Convert Temperature to Kelvin: T (K) = T (°C) + 273.15.
  • Solve for ΔG: Subtract the product of T and ΔS from ΔH.
Variables in Gibbs Free Energy Calculation
Variable Meaning Unit Typical Range
ΔG Gibbs Free Energy Change kJ/mol -500 to +500 kJ/mol
ΔH Enthalpy Change kJ/mol -1000 to +1000 kJ/mol
T Absolute Temperature Kelvin (K) 0 to 5000 K
ΔS Entropy Change J/mol·K -500 to +500 J/mol·K

Practical Examples (Real-World Use Cases)

Example 1: The Synthesis of Ammonia

Suppose you need to calculate delta g reaction using the following information for the Haber process at 298 K: ΔH = -92.2 kJ/mol and ΔS = -198.7 J/mol·K.

  • ΔH = -92.2 kJ/mol
  • T = 298 K
  • ΔS = -0.1987 kJ/mol·K
  • ΔG = -92.2 – (298 * -0.1987)
  • ΔG = -92.2 + 59.2 = -33.0 kJ/mol

The result is negative, meaning the reaction is spontaneous at room temperature.

Example 2: Evaporation of Water

Calculate ΔG for H2O(l) → H2O(g) at 25°C (298 K). Given ΔH = +44.0 kJ/mol and ΔS = +118.7 J/mol·K.

  • ΔG = 44.0 – (298 * 0.1187)
  • ΔG = 44.0 – 35.37 = +8.63 kJ/mol

Since ΔG is positive, water does not spontaneously boil at 25°C.

How to Use This calculate delta g reaction using the following information Calculator

  1. Enter Enthalpy (ΔH): Input the value from your data table or experiment. Ensure it is in kJ/mol.
  2. Input Entropy (ΔS): Note that most tables provide ΔS in Joules (J). Our calculator handles the conversion if you enter the J/mol·K value.
  3. Set the Temperature: Choose between Celsius or Kelvin. The tool automatically converts Celsius to Kelvin for the math.
  4. Analyze the Primary Result: Look at the large green (or red) box. A negative value indicates spontaneity.
  5. Check the Trend Chart: Observe how ΔG changes with temperature to find the “cross-over” point where a reaction becomes spontaneous.

Key Factors That Affect calculate delta g reaction using the following information Results

  • Magnitude of ΔH: Large exothermic reactions (negative ΔH) strongly favor spontaneity.
  • Sign of ΔS: Reactions that increase disorder (positive ΔS) are more likely to be spontaneous at high temperatures.
  • Absolute Temperature: High temperatures amplify the effect of the entropy term (-TΔS).
  • State Changes: Transitions from solid to liquid or liquid to gas significantly increase ΔS.
  • Concentration (Q): Under non-standard conditions, the reaction quotient affects the actual free energy.
  • Pressure: For gaseous reactions, changing the partial pressure shifts the entropy and thus the ΔG.

Frequently Asked Questions (FAQ)

What does a ΔG of zero mean?
It indicates the system is at chemical equilibrium; no net change occurs in the concentrations of reactants and products.

Can I calculate delta g reaction using the following information if I only have K?
Yes, using the formula ΔG° = -RT ln(K), where R is the gas constant and K is the equilibrium constant.

Why is temperature always in Kelvin?
Thermodynamic equations require an absolute scale where zero represents a total lack of thermal energy to prevent negative temperature values from flipping signs incorrectly.

Does a negative ΔG mean the reaction is fast?
No. ΔG only determines if the reaction is possible. The speed depends on the activation energy (kinetics).

How do I find ΔH and ΔS values?
These are typically found in standard thermodynamic property tables in chemistry textbooks or databases like NIST.

What is the difference between ΔG and ΔG°?
ΔG° is the free energy change under standard conditions (1 atm, 298 K, 1M concentrations), while ΔG is for any specific conditions.

What if ΔH and ΔS are both positive?
The reaction will be non-spontaneous at low temperatures but become spontaneous at high temperatures.

What if ΔH and ΔS are both negative?
The reaction will be spontaneous at low temperatures but become non-spontaneous as temperature increases.

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