Q10 Calculator

Work out Q10 from two rates measured at two temperatures.

Reads temperatures in °C or K and solves any one of the five boxes.

Updated October 12, 2026

CreatorFrank Zhao

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What this calculator is for

Two rates, two temperatures, one number. This page turns a pair of rate measurements into the Q10 temperature coefficient — the factor a reaction rate is multiplied by for every ten degree rise — and because the same equation reads backwards, it will also predict a rate you have not measured or recover a temperature you forgot to write down.

The five boxes are Temperature (T1), Temperature (T2), Reaction rate (R1), Reaction rate (R2) and Temperature coefficient (Q10). Fill in any four and the fifth is worked out as you type. There is nothing else to set up: no rate law, no units to pick for the rates, because a coefficient is defined purely from what you measured.

5 directions

Any four boxes determine the fifth

°C or K

Each temperature box takes either scale

1 equation

Q10 = (R₂/R₁) raised to 10/ΔT

The people who use it are usually in one of these situations: comparing an assay run on two different days, checking how far a warm incubation will push a reaction, turning field or bioreactor logging data into a number they can quote, or working through a kinetics problem where one of the five quantities is the unknown. Every figure in this guide came out of the calculator above, so you can reproduce any of them in a few keystrokes.

Quick start

Four numbers in, one number out. The path below is the shortest complete calculation, using a rate that doubles while the temperature climbs ten degrees.

Enter 20 as the first temperature

The temperature boxes open in Celsius, so type 20 into Temperature (T1). Readings below zero are accepted as long as they stay above absolute zero — a chilled sample at −5 °C is an ordinary measurement.

Enter 30 as the second temperature

Temperature (T2) has to sit above T1. The coefficient is defined over a rise, not a fall, and the page tells you straight away if the pair is the wrong way round.

Type 5 into Reaction rate (R1) and 10 into Reaction rate (R2)

The rates can be on any scale you like — per minute, per second, per hour — as long as both boxes are on the same one. Only their ratio enters the equation, and a ratio of quantities in different units is not one.

Read the coefficient that fills itself in

Temperature coefficient (Q10) is calculated for you, and it says exactly what you typed: the rate doubled over a ten degree rise.

Q_{10}

=

\left(\frac{10}{5}\right)^{\frac{10}{10}}

=

2

The calculated box is not locked

Only one box is solved at a time, and the four you typed keep their identity. Type over the calculated box and the page answers a different question instead of refusing you: type 3 over the calculated 2 and the first temperature is worked out again to keep the equation true, landing at about 23.69 °C for the numbers above. Ask a new question whenever you like — there is no form to clear first.

How the calculation works

One power law does all the work, and it is written the same way whichever box ends up being the answer.

Q_{10} = \left(\frac{R_2}{R_1}\right)^{\frac{10}{T_2 - T_1}}

The bracket is the ratio of the two measured rates. The exponent rescales that ratio to a single ten-degree step, so a gap of twenty degrees contributes an exponent of ½ and a gap of five degrees an exponent of 2. Written with logarithms — the form the reverse directions fall out of — the same statement is:

Q_{10} = \exp\left(\frac{10}{T_2 - T_1}\,\ln\frac{R_2}{R_1}\right)

Q_{10}

Temperature coefficient: the multiplier for a ten degree rise

R_1

Reaction rate measured at temperature T1

R_2

Reaction rate measured at temperature T2

T_1

First measurement temperature, typed in °C or K

T_2

Second measurement temperature, above T1

\Delta T = T_2 - T_1

The gap between the two readings — identical in °C and in K

Running it backwards

Because the relation contains exactly five quantities, any one of them can be the answer. Predicting a rate at a new temperature uses the exponent on the rate you already have:

R_2 = R_1 \times Q_{10}^{\,\Delta T / 10}

Recovering a temperature divides the logarithm of the rate ratio by the logarithm of the coefficient:

\Delta T

=

\frac{10\,\ln(R_2 / R_1)}{\ln Q_{10}}

Two inputs make the equation impossible rather than merely difficult: identical temperatures leave 10/(T2 − T1) with nothing to divide by, so the coefficient stays blank and explains why instead of inventing an infinity.

Both temperatures, either scale

The equation only ever consumes the gap between the readings, and a Celsius reading carries the same 273.15 offset as the kelvin reading beside it, so the offset cancels inside the subtraction:

50 - 30 = 20\ \mathrm{^\circ C}

=

323.15 - 303.15 = 20\ \mathrm{K}

You can even leave one box in Celsius and switch the other to kelvins; the gap comes out the same either way. The rates need no conversion at all, because a ratio of two numbers in the same units has no units left.

What different values mean

The table below is arithmetic, not a survey: it is simply what each coefficient does to a rate over one and two ten-degree steps.

Q10	After +10 °C	After +20 °C	What it says
0.5	× 0.5	× 0.25	Rate halves each ten degrees
1	× 1	× 1	Rate ignores temperature
1.5	× 1.5	× 2.25	Rate grows by half each ten degrees
2	× 2	× 4	Rate doubles each ten degrees
3	× 3	× 9	Rate triples each ten degrees

A value of 1 is the flat case: the two rates came out equal, so the measurement carries no temperature information at all. A value below 1 is the honest answer to a rate that fell while the temperature rose, which is worth pausing on — see the questions at the end before you decide it is a mistake.

The same dependence, written as an energy

Temperature dependence also has a deeper form, the Arrhenius equation $k = A\,e^{-E_{\mathrm a}/RT}$ , and the two descriptions can be converted into each other. Combining the definition above with it cancels the gap and leaves the coefficient depending only on the activation energy and the temperature level:

Q_{10} = \exp\left(\frac{10\,E_{\mathrm a}}{R\,T_1 T_2}\right)

Run it the other way for the worked example above: a coefficient of 2 measured between 303.15 K and 323.15 K corresponds to an activation energy of about 56.5 kJ/mol, using the molar gas constant R = 8.314 J·mol⁻¹·K⁻¹ tabulated by NIST. Notice what the denominator does: with an activation energy of 50 kJ/mol the same reaction reports a coefficient near 2.10 between 280 K and 290 K, but only about 1.66 between 340 K and 350 K. A coefficient belongs to the window it was measured in, and this is why.

When what you need is the rate constant, the frequency factor, the activation energy or the temperature itself rather than a single coefficient, the Arrhenius equation calculator solves that equation in any direction.

Common scenarios

Three questions the five boxes answer, each with the numbers the calculator actually produced.

Scenario 1

Two measurements become a coefficient

An assay ran at 25 °C with a rate of 4, and the same assay at 35 °C with a rate of 9. Four boxes in, the coefficient comes back at 2.25 — the rate grows by a factor of 2.25 for every ten degrees across that range.

Q_{10} = \left(\tfrac{9}{4}\right)^{10/10}

=

2.25

Use it: a run that sat 3 °C warm will travel about 1.28 times faster than the 35 °C rate, because $2.25^{0.3} \approx 1.28$ .

Measured pair → Q10

Scenario 2

Predict a rate you did not measure

Continuing the quick start: twenty degrees at a rate of 5, coefficient 2. What should the rate be at 45 °C? Leave R2 empty and the exponent does the work.

R_2 = 5 \times 2^{25/10}

=

28.2843

Use it: the same relation predicts a rate at any temperature your window can defend — check the tips below before you push it far.

Coefficient + gap → new rate

Scenario 3

Recover a temperature you did not log

You have a rate of 5 and a rate of 10, you know the coefficient is 2, and the first reading was 30 °C — but the second log line is missing. Type the four you have and the temperature solves itself.

T_2 = 30 + \frac{10\,\ln(10/5)}{\ln 2}

=

40\ \mathrm{^\circ C}

Use it: the same back-solve fills a gap in a temperature log from two rates you already have.

Rate ratio → missing temperature

If your rates are not numbers yet

This page starts from two rates that already exist. If all you have is concentrations and a rate law, the rate constant calculator works out the rate for you, and the number it gives can be typed straight into R1 or R2 here.

Tips & best practices

The equation is trivial; the measurement is where coefficients go wrong. These five habits separate a number you can quote from one you have to apologise for.

Widen the gap between the two temperaturesPrecision

The exponent is $10/\Delta T$ , so a gap that is misread by one degree moves the answer a long way. The same rate ratio of 2 gives 2 over a ten degree gap, 2.37841 if the gap was really eight degrees, and 1.7818 if it was twelve — the two extremes are a third apart. Ten degrees is a workable floor; twenty is comfortable.

Keep both rates on the same scaleUnits

Only the ratio $R_2 / R_1$ enters the equation, so 5 and 10 agree, while a rate written per minute beside a rate written per second quietly multiplies your answer by 60.

Measure both rates the same wayMethod

Same substrate batch, same assay duration, same instrument settings. Any difference that is not temperature gets folded into the coefficient and comes back out looking like temperature sensitivity.

Treat a coefficient near 1 with care when solving backwardsBack-solving

Recovering a gap divides by $\ln Q_{10}$ , which is nearly zero there: a ratio of 2 implies a 10 degree gap at $Q_{10} = 2$ , but about 142 degrees at $Q_{10} = 1.05$ . A rounding nudge in the inputs can swing the recovered temperature by tens of degrees.

Do not carry the number far outside its windowExtrapolation

One coefficient describes one interval. For a wider range, measure a second pair and check that it agrees, or work from the Arrhenius form quoted above.

Limitations & disclaimers

This is a two-point model. It compresses everything about a temperature response into a single multiplier, and that compression is where it stops being honest.

Where the model stops being exact

One window, one coefficient
The value describes the pair of temperatures you typed in. Push it well beyond that range and the error compounds with every degree, because a real response is rarely a clean geometric progression.
Systems that change state break it
A catalyst that is losing activity, a solution approaching its boiling point, a biological component past the temperature where it still functions — the rate stops climbing smoothly, and no single coefficient describes both sides of that turn.
The inputs have physical boundaries
Every box must be a positive value, temperatures must sit above absolute zero, and T2 must be greater than T1. Identical temperatures leave the gap undefined, and a coefficient of 1 carries no temperature information, so those cases are reported rather than answered.
Measurement error passes straight through
The coefficient is built from two rates and two temperatures; whatever uncertainty those four numbers carry is amplified by the exponent, most sharply when the gap is narrow.
Purpose of this tool
Results are for education, experiment planning and quick comparison. Anything that sets a process temperature, a shelf-life claim or a safety limit should be checked against measurements taken across your actual operating range.

When a single number is not enough

A coefficient summarises a window; it does not model the chemistry inside it. If you need the rate at a series of temperatures, or an activation energy you can defend, take several measurements and fit the Arrhenius form instead of extrapolating one Q10 — and for work that carries process or safety consequences, have the fit reviewed against real data by someone qualified to sign off on it.

Frequently asked questions

Why is my coefficient below 1?

Because the rate fell while the temperature rose. Ten down to five over a ten degree rise gives exactly 0.5, and the calculator will happily show it. Two explanations are common: the pair of rates is transcribed or scaled wrongly (check that both really are on the same units), or the system has passed the point where more heat still helps, so you are measuring decline rather than temperature sensitivity. Either way the arithmetic is right — it is the interpretation that needs a second look.

Does it matter whether I type Celsius or kelvins?

No. The two readings are compared on the kelvin scale, and since a Celsius reading and a kelvin reading of the same temperature differ by a fixed 273.15, that offset cancels when the two are subtracted:

50 - 30 = 20\ \mathrm{^\circ C}

=

323.15 - 303.15

=

20\ \mathrm{K}

Each box converts on its own, so you can even leave one in Celsius and read the other in kelvins — the gap, and therefore the coefficient, will not move.

Why does the second temperature box turn red?

Because T2 has to sit above T1. The coefficient is defined over a temperature rise, and a number that solves the equation to a second temperature below the first is reported rather than hidden: enter T1 = 30 °C, rates of 10 and 5, and a coefficient of 2 and the page computes 20 °C, leaves it on screen, and marks it with the message "Temperature T2 should be greater than temperature T1." The same message appears when the two temperatures are equal, since a gap of zero cannot produce a coefficient at all.

Can it find a temperature for me when the coefficient is 1?

No, and nothing could. Unfolding a temperature divides by $\ln Q_{10}$ , and $\ln 1 = 0$ . A coefficient of 1 says the rate did not change at all, which is equally consistent with every temperature gap — there is no number to recover, so the box stays blank and says why.

I measured 20–30 °C. Can I use that coefficient at 60 °C?

Not without checking it. The coefficient you measured is an average over the window it came from, and the Arrhenius form in the method section shows why it drifts: for a fixed activation energy the same reaction reports a smaller coefficient as the temperature level rises. Measure a second pair closer to 60 °C and compare the two — if they agree, extrapolating a little is reasonable; if they do not, the coefficient is telling you the response is curving, and a single number is the wrong tool for that range.

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