Calculate Gibbs free energy, predict reaction spontaneity, analyze temperature dependence, and explore equilibrium relationships using ΔG = ΔH - TΔS for chemistry problems and exam prep.
Select an operation and enter thermodynamic values to calculate Gibbs free energy, analyze spontaneity, and visualize temperature dependencies.
Gibbs free energy (ΔG) is the ultimate arbiter of chemical spontaneity—it tells you whether a reaction will proceed forward under given conditions. The relationship ΔG = ΔH - TΔS elegantly combines two competing factors: enthalpy (ΔH, the heat absorbed or released) and entropy (ΔS, the change in disorder or energy dispersal), both modulated by absolute temperature (T in Kelvin). When ΔG is negative, the reaction is thermodynamically favorable and will tend to proceed forward. When ΔG is positive, the reaction is unfavorable as written, and the reverse direction is preferred.
In everyday chemistry, this framework explains why ice melts at room temperature (entropy-driven: positive ΔH but larger positive TΔS term makes ΔG negative), why combustion reactions release heat and are spontaneous (exothermic and entropy-increasing), and why some endothermic processes like dissolving ammonium nitrate can still proceed (entropy gain compensates for heat absorption). Temperature plays a pivotal role: a reaction unfavorable at low temperature might become spontaneous when heated, or vice versa.
This calculator helps students and learners solve thermodynamics problems quickly and accurately, whether you're checking homework answers, preparing for chemistry exams, analyzing equilibrium scenarios, or exploring how temperature shifts spontaneity. Understanding these relationships builds intuition for more advanced topics like chemical equilibria, phase transitions, electrochemistry, and biochemical energetics. Remember: thermodynamics tells you if a reaction can happen, not how fast—that's the realm of kinetics.
Step 1: Select the calculation type based on your problem. The most common task is computing ΔG from ΔH, ΔS, and T using the Gibbs-Helmholtz equation. Other modules handle equilibrium constants (relating ΔG° to K), reaction quotients (non-standard conditions), van't Hoff analysis (temperature dependence of K), and temperature-dependent spontaneity plots.
Step 2: Enter known values with correct units. ΔH is typically given in kJ/mol or J/mol. ΔS is usually in J/(mol·K)—note the different energy units. Temperature must be in Kelvin (K = °C + 273.15). Ensure ΔH and ΔS use compatible energy units before calculation. If ΔH is in kJ/mol and ΔS in J/(mol·K), convert ΔH to J/mol by multiplying by 1000, or convert ΔS to kJ/(mol·K) by dividing by 1000.
Step 3: Understand sign conventions. ΔH < 0 means exothermic (heat released, like combustion). ΔH > 0 means endothermic (heat absorbed, like melting ice). ΔS > 0 means entropy increases (more disorder, like gas formation). ΔS < 0 means entropy decreases (more order, like freezing). These signs, combined with temperature, determine the sign and magnitude of ΔG.
Step 4: Interpret results. ΔG < 0 indicates a spontaneous (thermodynamically favorable) process under the specified conditions. ΔG > 0 means non-spontaneous (the reverse reaction is favored). ΔG ≈ 0 suggests the system is at or near equilibrium. Use the calculator's visualizations to explore how ΔG changes with temperature, revealing crossover points where spontaneity flips. Copy results for inclusion in lab reports or problem sets.
This is the master equation for spontaneity. ΔG (Gibbs free energy change) combines enthalpy ΔH and entropy ΔS, weighted by temperature T (in Kelvin). The equation shows that even if ΔH is unfavorable (positive, endothermic), a sufficiently large positive ΔS can make the TΔS term dominate at high temperature, yielding negative ΔG (spontaneous). Conversely, exothermic reactions (ΔH < 0) with positive ΔS are spontaneous at all temperatures.
Example: Reaction with ΔH = -100 kJ/mol, ΔS = -200 J/(mol·K) at T = 298 K:
Convert ΔS: -200 J/(mol·K) = -0.200 kJ/(mol·K)
ΔG = ΔH - TΔS = -100 - (298)(-0.200) = -100 + 59.6 = -40.4 kJ/mol
ΔG < 0 → reaction is spontaneous at 298 K despite negative entropy change
Standard Gibbs free energy change ΔG° (at 1 bar, 298 K, 1 M concentrations) relates to the equilibrium constant K. If K > 1, products are favored at equilibrium (ΔG° < 0). If K < 1, reactants dominate (ΔG° > 0). R is the gas constant (8.314 J/(mol·K)). Large negative ΔG° means K is very large (reaction goes nearly to completion). This equation connects thermodynamics to equilibrium quantitatively.
When conditions aren't standard, use the reaction quotient Q (calculated like K but with current concentrations/pressures). If Q < K, ΔG < 0 and the forward reaction proceeds. If Q > K, ΔG > 0 and the reverse reaction proceeds. If Q = K, ΔG = 0 and the system is at equilibrium. This equation predicts reaction direction for any set of concentrations.
The van't Hoff equation describes how equilibrium constant K varies with temperature. Plotting ln K vs 1/T yields a straight line with slope -ΔH°/R and intercept ΔS°/R, allowing experimental determination of ΔH° and ΔS° from K measurements at multiple temperatures. Exothermic reactions (ΔH° < 0) have K decreasing as T increases. Endothermic reactions (ΔH° > 0) have K increasing with T.
There are four scenarios based on ΔH and ΔS signs: (1) ΔH < 0, ΔS > 0 → always spontaneous (ΔG always negative). (2) ΔH > 0, ΔS < 0 → never spontaneous (ΔG always positive). (3) ΔH < 0, ΔS < 0 → spontaneous at low T (enthalpy-driven), non-spontaneous at high T when TΔS term dominates. (4) ΔH > 0, ΔS > 0 → non-spontaneous at low T, spontaneous at high T (entropy-driven). The crossover temperature is T_eq = ΔH/ΔS where ΔG = 0.
Typical units: ΔG and ΔH in kJ/mol (or J/mol), ΔS in J/(mol·K), T in Kelvin, R = 8.314 J/(mol·K) or 0.008314 kJ/(mol·K). Critical conversion: if ΔH is in kJ/mol and ΔS is in J/(mol·K), convert one to match the other. For temperature: K = °C + 273.15. Always use Kelvin for thermodynamic calculations to avoid sign errors.
Students use this calculator to verify thermodynamics problem solutions, check sign conventions, and build intuition for how ΔH, ΔS, and T interact. Instead of memorizing tables, you develop understanding by seeing how changing temperature affects ΔG, or how different combinations of ΔH and ΔS signs lead to always-spontaneous, never-spontaneous, or temperature-dependent scenarios.
A reaction might be non-spontaneous at room temperature but become favorable when heated. For example, thermal decomposition of calcium carbonate (CaCO₃ → CaO + CO₂) is endothermic (ΔH > 0) with positive ΔS (gas formation). At low T, ΔG > 0 (non-spontaneous), but above ~1100 K, TΔS exceeds ΔH, making ΔG < 0 (spontaneous). This calculator helps find that crossover temperature.
Some reactions are driven by entropy (like dissolving a salt where ΔH > 0 but ΔS > 0), while others are driven by enthalpy (like combustion where ΔH < 0 dominates). Use the calculator to explore which term (ΔH or TΔS) dominates at various temperatures. This clarifies why ice melts at room temperature (entropy-driven despite being endothermic) or why NH₄NO₃ dissolves endothermically (entropy increase compensates for heat absorption).
Given ΔG° for a reaction, calculate K using ΔG° = -RT ln K. A ΔG° of -10 kJ/mol at 298 K gives K ≈ 56 (products strongly favored). A ΔG° of +10 kJ/mol gives K ≈ 0.018 (reactants favored). This helps predict whether a reaction will produce significant products or barely proceed. Conversely, from measured K, you can back-calculate ΔG° to understand the thermodynamic driving force.
Phase changes (melting, boiling, sublimation) are equilibrium processes where ΔG = 0 at the transition temperature. For water boiling at 100°C (373 K), ΔG = 0, so ΔH_vap = T·ΔS_vap. Use the calculator to estimate boiling points by finding where ΔG crosses zero. This is conceptual—actual lab determination requires precise ΔH and ΔS values—but it illustrates how thermodynamics governs phase equilibria.
In electrochemical cells, ΔG relates to cell potential E via ΔG = -nFE (n = moles of electrons, F = Faraday constant). A spontaneous redox reaction (ΔG < 0) produces positive cell voltage (galvanic cell). This calculator helps understand the thermodynamic foundation of batteries and electrolysis. While electrochemistry has its own tools, knowing ΔG provides the energy perspective.
If you've measured K at several temperatures in a lab experiment, use the van't Hoff module to fit ln K vs 1/T. The slope and intercept give ΔH° and ΔS° with uncertainties (R² indicates fit quality). This is a common physical chemistry lab exercise. The calculator streamlines the analysis, letting you focus on interpreting the thermodynamic parameters rather than manual curve fitting.
Compare two reactions: Reaction A has ΔH = -50 kJ/mol, ΔS = -100 J/(mol·K); Reaction B has ΔH = -30 kJ/mol, ΔS = +50 J/(mol·K). At 298 K, which is more spontaneous? Calculate both ΔG values. Reaction A: ΔG = -50 - 298(-0.100) = -50 + 29.8 = -20.2 kJ/mol. Reaction B: ΔG = -30 - 298(0.050) = -30 - 14.9 = -44.9 kJ/mol. Reaction B is more spontaneous despite smaller ΔH because entropy favors it.
Using temperature in Celsius instead of Kelvin
The equation ΔG = ΔH - TΔS requires absolute temperature in Kelvin. Using °C directly gives wildly incorrect results. Always convert: K = °C + 273.15. At 25°C, use 298.15 K, not 25. Forgetting this is the #1 error in thermodynamics problems, leading to wrong signs and magnitudes for ΔG.
Mixing energy units for ΔH and ΔS
ΔH is often in kJ/mol while ΔS is in J/(mol·K). Before using ΔG = ΔH - TΔS, convert to matching units: either ΔH to J/mol (multiply by 1000) or ΔS to kJ/(mol·K) (divide by 1000). Failing to convert gives ΔG off by a factor of 1000. For example, if ΔH = -50 kJ/mol and ΔS = 100 J/(mol·K), convert ΔS to 0.100 kJ/(mol·K) before calculating.
Confusing spontaneity with reaction rate (kinetics vs thermodynamics)
ΔG < 0 means thermodynamically favorable, not necessarily fast. Diamond converting to graphite is spontaneous (ΔG < 0) but happens so slowly that diamonds are effectively stable. Conversely, ΔG > 0 doesn't mean the reaction can't happen at all—it means it won't proceed significantly under equilibrium conditions. Thermodynamics says "can it happen?" Kinetics says "how fast?"
Assuming ΔH and ΔS are constant over very large temperature ranges
The simple ΔG = ΔH - TΔS assumes ΔH and ΔS are temperature-independent, which is an approximation. Over small temperature ranges (e.g., 200-400 K), it's usually fine. Over very large ranges (e.g., 100-1000 K), ΔH and ΔS actually change due to heat capacity effects. For introductory courses, the constant-ΔH-and-ΔS approximation is standard. For precision work, use the Kirchhoff equations to account for temperature dependence.
Confusing standard ΔG° with non-standard ΔG
ΔG° is defined at standard conditions (1 bar, 298 K, 1 M). ΔG applies to any conditions via ΔG = ΔG° + RT ln Q. Don't use ΔG° to predict spontaneity at non-standard concentrations—use the full equation with Q. For example, a reaction with ΔG° = +5 kJ/mol (non-spontaneous under standard conditions) can still be spontaneous if Q << 1 (very low product concentrations).
Treating ΔG as a directly measurable quantity
You can't "measure ΔG" in a single experiment the way you measure temperature or mass. ΔG is calculated from measurable quantities: ΔH (calorimetry), ΔS (heat capacity measurements), or K (equilibrium measurements). When problems say "given ΔG," they mean it's been calculated from other data. Understanding this prevents conceptual confusion about what thermodynamic quantities represent.
Misinterpreting the sign of ΔS for a process
Positive ΔS means entropy increases (more disorder): solid → liquid → gas, or reactants → more product molecules, or dissolving. Negative ΔS means entropy decreases: gas → liquid → solid, or more molecules → fewer, or crystallizing from solution. Don't confuse ΔS sign with ΔH sign. A reaction can be exothermic (ΔH < 0) with either positive or negative ΔS depending on the molecular disorder change.
Forgetting the negative sign in ΔG° = -RT ln K
The equation is ΔG° = -RT ln K, not +RT ln K. The negative sign ensures that K > 1 (products favored) corresponds to ΔG° < 0 (spontaneous). If you drop the negative sign, you'll get the wrong sign for ΔG°, leading to incorrect conclusions about equilibrium favorability. Always include it in calculations and double-check your algebra.
Using natural log (ln) vs common log (log₁₀) incorrectly
Thermodynamic equations use natural logarithm (ln, base e) not common logarithm (log₁₀, base 10). ΔG° = -RT ln K uses ln, not log₁₀. On calculators, ensure you press "ln" not "log". Mixing these up gives wrong answers by a factor of ln(10) ≈ 2.303. If you see log₁₀ in older texts, convert: ln(x) = 2.303 log₁₀(x).
Not checking if results are physically reasonable
After calculating ΔG, verify it makes sense. For strongly exothermic reactions with positive ΔS, expect very negative ΔG. For endothermic reactions with negative ΔS, expect positive ΔG. If you get ΔG = -10,000 kJ/mol for a simple acid-base reaction, something went wrong (likely unit conversion). Dimensional analysis and sanity checks catch most errors.
Explore the four sign combinations systematically
Master the four ΔH/ΔS scenarios: (1) ΔH < 0, ΔS > 0 → always spontaneous (exothermic + disorder increase). (2) ΔH > 0, ΔS < 0 → never spontaneous (endothermic + disorder decrease). (3) ΔH < 0, ΔS < 0 → spontaneous at low T (enthalpy-driven). (4) ΔH > 0, ΔS > 0 → spontaneous at high T (entropy-driven). Use the calculator to visualize how ΔG changes with T for each scenario and find crossover temperatures.
Use temperature-dependence plots to predict phase transitions
For phase changes (solid ↔ liquid ↔ gas), ΔG = 0 at the transition temperature. Plot ΔG vs T using known ΔH and ΔS values for the transition. The temperature where the line crosses zero is the equilibrium transition temperature (melting point, boiling point). This connects thermodynamics to everyday phenomena like water freezing at 273 K and boiling at 373 K at 1 atm.
Connect ΔG to equilibrium position quantitatively
ΔG° relates to K via ΔG° = -RT ln K. For K = 1 (equal products and reactants at equilibrium), ΔG° = 0. For K = 100, ΔG° ≈ -11.4 kJ/mol at 298 K. For K = 0.01, ΔG° ≈ +11.4 kJ/mol. Plot ln K vs ΔG° to see the exponential relationship. This helps understand why small changes in ΔG° lead to large changes in K (and thus equilibrium position).
Use van't Hoff analysis to extract ΔH° and ΔS° from experimental K values
If you have K measured at multiple temperatures, plot ln K vs 1/T (van't Hoff plot). The slope is -ΔH°/R and intercept is ΔS°/R. This graphical method is robust: even if individual K measurements have scatter, linear regression smooths errors. Check R² to assess linearity. Deviations from linearity suggest ΔH° and ΔS° are temperature-dependent or experimental issues.
Understand the relationship between ΔG and cell potential in electrochemistry
For redox reactions, ΔG = -nFE where n is moles of electrons transferred and F = 96485 C/mol (Faraday constant). A spontaneous redox reaction (ΔG < 0) has E > 0 (galvanic cell). Use this to connect thermodynamics to batteries and electrolysis. Calculate E from ΔG° or vice versa. This dual perspective (ΔG and E) enriches understanding of redox processes.
Explore coupled reactions and Le Châtelier's principle thermodynamically
Two reactions can be coupled: if reaction A is non-spontaneous (ΔG_A > 0) but reaction B is highly spontaneous (ΔG_B << 0), the overall process A + B can have ΔG_total < 0. This is fundamental in biochemistry (ATP hydrolysis drives unfavorable reactions). Use the calculator to explore how adding a favorable reaction shifts the overall thermodynamic balance.
Distinguish between ΔG° (standard conditions) and ΔG (actual conditions)
ΔG° tells you about equilibrium position (via K). ΔG tells you the driving force right now at current concentrations (via Q). Even if ΔG° > 0 (K < 1, reactants favored at equilibrium), ΔG can be < 0 if Q < K (low product concentrations). This explains why non-spontaneous reactions can proceed forward initially before reaching equilibrium. Use both ΔG° and ΔG for complete analysis.
Use this calculator as a quick-check tool for textbook and exam problems
Before submitting homework or exams, verify your hand calculations with the calculator. If you calculated ΔG = -25 kJ/mol but the calculator gives -52 kJ/mol, recheck your work. Common errors: forgot to convert ΔS units, used °C instead of K, dropped a negative sign. This quality check improves accuracy and builds confidence in thermodynamic problem-solving.
Connect entropy changes to molecular-level disorder intuitively
ΔS > 0 often corresponds to: gas formation from solids/liquids (CO₂ release), dissolving a solid (ions dispersing), or more product molecules than reactants (2 molecules → 3 molecules). ΔS < 0 corresponds to opposite processes. Visualizing molecules spreading out (positive ΔS) or condensing (negative ΔS) helps predict ΔS sign before calculating, catching errors early.
Master the connection between thermodynamics and Le Châtelier's principle
Le Châtelier's principle (adding product shifts left, adding reactant shifts right) is thermodynamics in disguise. Adding product increases Q; if Q > K, then ΔG > 0 and the reverse reaction proceeds until Q = K again. Heating an endothermic reaction (ΔH > 0) favors products because increased T makes TΔS more positive (if ΔS > 0), shifting equilibrium right. Use thermodynamics to quantify Le Châtelier predictions.
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