Calculate entropy change (ΔS) from heat transfer, heating or cooling, ideal gas expansion, phase changes, or standard reaction entropies, plus Shannon entropy in bits.
Entropy Formula
Entropy change is always defined by the heat moved along a reversible path divided by the absolute temperature at which it moves. Every mode below is a special case of that one definition.
- dS = entropy change of the system (J/K)
- Q = heat transferred to the system (J), positive when heat is absorbed
- T = absolute temperature during the transfer (K)
Use this mode when the temperature does not move while the heat flows, such as a reservoir at fixed temperature. If you also supply the surroundings temperature, the calculator returns dS surroundings = -Q/Tsurr and the total, which is the quantity the second law constrains.
- C = total heat capacity of the sample (J/K), equal to n*Cp, m*c, or a value you enter directly
- T1 = starting absolute temperature (K)
- T2 = ending absolute temperature (K)
This is the heating and cooling mode. The temperature moves during the process, so the heat is integrated over the path and a logarithm appears instead of a simple division.
- n = moles of ideal gas
- R = gas constant, 8.314462618 J/mol*K
- V1, V2 = initial and final volume (any consistent unit)
- P1, P2 = initial and final pressure (any consistent unit)
This is the isothermal ideal gas mode. Only the ratio matters, so the units cancel. Volume and pressure inputs are never shown at the same time because they are two routes to the same number.
- dHtrans = enthalpy of fusion, vaporization, or sublimation (J/mol)
- T = transition temperature, the melting or boiling point (K)
This is the phase change mode. Temperature is pinned at the transition point, so the full latent heat converts to entropy. Choosing the reverse direction flips the sign for freezing, condensation, or deposition. For vaporization the calculator also reports the molar value next to Trouton's rule.
- S = standard molar entropy of each species (J/mol*K), each multiplied by its balanced coefficient
- dG = dH - T*dS, the optional spontaneity check
- Tcross = dH/dS, the temperature where the sign of dG flips
This is the reaction mode. Standard entropies are absolute values referenced to a perfect crystal at 0 K, so they are never negative for a real substance even though the reaction total often is.
- p_i = probability of outcome i, taken from your list, your counts, or the character frequencies of your text
- b = log base, 2 for bits, e for nats, 10 for hartleys
- Hmax = log_b(k) for k outcomes, used for the efficiency and redundancy figures
This is the information entropy mode, which shares the same mathematical form as the statistical definition of thermodynamic entropy but reports bits rather than J/K.
Reference Values for Checking Your Result
Trouton's rule says the molar entropy of vaporization at the normal boiling point sits near 88 J/mol*K for most liquids, because a mole of any liquid gains roughly the same disorder when it becomes a mole of gas. The interesting part is the deviation. Liquids that are already organized by hydrogen bonding start from a lower entropy, so they gain more on boiling and land well above 88. Liquids whose vapor is partly ordered or that boil near absolute zero land below it. The calculator prints this deviation for you in vaporization mode.
| Liquid | dHvap (kJ/mol) | Tb (K) | dSvap (J/mol*K) | vs 88 |
|---|---|---|---|---|
| Helium | 0.08 | 4.2 | 19 | -69 |
| Hydrogen | 0.90 | 20.3 | 44 | -44 |
| Hexane | 28.85 | 341.9 | 84 | -4 |
| Carbon tetrachloride | 29.82 | 349.9 | 85 | -3 |
| Diethyl ether | 26.5 | 307.6 | 86 | -2 |
| Benzene | 30.72 | 353.2 | 87 | -1 |
| Acetone | 29.1 | 329.2 | 88 | 0 |
| Mercury | 59.1 | 629.9 | 94 | +6 |
| Ammonia | 23.35 | 239.8 | 97 | +9 |
| Methanol | 35.2 | 337.7 | 104 | +16 |
| Water | 40.65 | 373.15 | 109 | +21 |
| Ethanol | 38.6 | 351.4 | 110 | +22 |
The next table gives the size you should expect for a result before you trust it. If your answer is off by an order of magnitude from the matching row, the usual causes are Celsius left in place of Kelvin, kJ mixed with J, or a molar quantity reported as a total.
| Process | Typical dS | Sign driver |
|---|---|---|
| Melting one mole of a metal | +8 to +14 J/mol*K | Lattice positions unlock |
| Melting one mole of a molecular solid | +20 to +60 J/mol*K | Rotation and orientation unlock too |
| Boiling one mole of liquid | +85 to +110 J/mol*K | Roughly a thousandfold volume gain |
| Doubling the volume of one mole of gas | +5.76 J/K | R*ln(2), independent of the gas |
| Heating one mole of a monatomic gas 300 to 600 K | +14.4 J/K at constant P | Cp*ln(2) with Cp = 20.8 |
| Reaction that creates one extra mole of gas | +100 to +200 J/mol*K | Gas moles dominate the total |
| Reaction with no change in gas moles | -30 to +30 J/mol*K | Only structural differences remain |
| Reaction that consumes gas into a solid | -150 to -350 J/mol*K | Gas moles removed |
Example Problems
Example 1: vaporizing water. You boil 0.500 mol of water at 373.15 K. The enthalpy of vaporization is 40.65 kJ/mol. Select the phase change mode, set the transition to vaporization and the direction to forward, then enter n = 0.5, dH = 40.65, and T = 373.15.
dS = n*dHtrans/T = 0.500 * 40,650 / 373.15 = 54.5 J/K
The molar figure is 108.9 J/mol*K, about 21 J/mol*K above Trouton's rule. That gap is the hydrogen bonding in liquid water showing up as a number: the liquid was unusually ordered to begin with, so boiling it buys more entropy than boiling a comparable nonpolar liquid.
Example 2: heating water in a kettle. You heat 250 g of water from 25 degrees Celsius to 80 degrees Celsius. Specific heat is 4.184 J/g*K. Select the heating or cooling mode, set the basis to specific heat, set the temperature unit to Celsius, then enter c = 4.184, m = 250, T1 = 25, T2 = 80.
C = 250 * 4.184 = 1046 J/K, and T converts to 298.15 K and 353.15 K.
dS = 1046 * ln(353.15/298.15) = 1046 * 0.1692 = 177.0 J/K
The heat added is 1046 * 55 = 57.5 kJ, so a careless Q/T using the final temperature would give 163 J/K, about 8 percent low. That gap is exactly why the logarithm is there.
FAQ
When can you use Q/T instead of the logarithm?
Only when the temperature of the system holds constant while the heat moves, which in practice means a phase change, an isothermal expansion, or contact with a reservoir large enough that its temperature does not shift. As soon as the sample warms or cools, each joule enters at a different temperature and you need dS = C*ln(T2/T1). The error from ignoring this grows with the temperature ratio: over a 5 percent temperature change it is under 0.1 percent, but heating from 298 K to 800 K the logarithm gives 0.988*C while dividing the heat by the average temperature gives 0.914*C, a 7 percent miss.
Can entropy change be negative?
Yes, for a system. Freezing water, compressing a gas, and reactions that pull gas molecules into a solid all give negative values, and the calculator will return them. The second law constrains the total, not the system. Whenever a system loses entropy the surroundings must gain at least as much, which is why the heat transfer mode offers a surroundings temperature: it returns dS system plus dS surroundings, and that sum has to be zero or positive for the process to happen on its own.
Is Shannon entropy the same thing as thermodynamic entropy?
They are the same formula measuring different things. Thermodynamic entropy in the statistical picture is S = k*ln(W), the log of the number of microstates, and Shannon entropy H = -sum(p*log(p)) reduces to exactly that when every microstate is equally likely. The link is numerical: one bit of information entropy corresponds to k*ln(2), or about 9.57e-24 J/K. So a result of 2 bits per symbol and a result of 2 J/K are not comparable quantities, and the calculator keeps the two modes separate for that reason. Use the information mode for compression limits, decision tree splits, and password strength, and the thermodynamic modes for heat and reactions.