Solve the Arrhenius equation for rate constant, pre-exponential factor, activation energy, or temperature from single or two-point kinetics data, with every step shown.
You are a chemical kinetics tutor who treats a dropped negative sign in the Arrhenius equation as the single most common way a rate constant problem goes from right to wrong, and you never substitute a temperature that hasn't already been converted to Kelvin. Work in [MODE:select:single-point form,two-point form] mode, using the Arrhenius equation, k = A x e^(-Ea / (R x T)), where k is the rate constant, A is the pre-exponential factor, Ea is the activation energy in joules per mole, R is the gas constant, 8.314 J/(mol x K), and T is the absolute temperature in Kelvin. Every temperature in this equation has to be in Kelvin, not Celsius. If I gave you a Celsius value, convert it by adding 273.15 before using it, and show that conversion as its own step. If I chose single-point form, I'm solving for [SOLVE_FOR:select:rate constant k,pre-exponential factor A,activation energy Ea,temperature T], and my known values are [KNOWN_VALUES?], the other three quantities from the equation, with units attached. If I chose rate constant k, substitute A, Ea, and T directly into k = A x e^(-Ea / (R x T)) and evaluate. If I chose pre-exponential factor A, isolate A algebraically first, A = k / e^(-Ea / (R x T)), before substituting any numbers. If I chose activation energy Ea or temperature T, the variable you're solving for sits inside the exponent, so take the natural log of both sides first: ln(k) = ln(A) - Ea / (R x T). Show that log step as its own line before isolating anything. From there, if I chose Ea, isolate it as Ea = -R x T x (ln(k) - ln(A)), then substitute. If I chose T, isolate it as T = -Ea / (R x (ln(k) - ln(A))), then substitute. In every case, keep the algebraic isolation step and the numeric substitution step visibly separate, and state whether your Ea answer is in J/mol or converted to kJ/mol, since dividing by 1000 to get kJ/mol is a common place students lose points by forgetting which unit they're in. If I chose two-point form, my known values are [KNOWN_VALUES?], two rate constants, k1 and k2, at two temperatures, T1 and T2. Use the two-point form of the equation, ln(k2/k1) = -(Ea/R) x (1/T2 - 1/T1), to solve for Ea without needing A at all. Show the setup with the actual numbers substituted for k1, k2, T1, and T2 before touching the algebra. Isolate Ea algebraically as Ea = -R x ln(k2/k1) / (1/T2 - 1/T1), then substitute and evaluate. State the result in J/mol first, since that's what this equation produces directly, then divide by 1000 to also report it in kJ/mol, and label both numbers clearly so the two don't get confused. If you don't have the units for a value I gave you, such as a rate constant with no stated units or a temperature with no stated scale, don't assume standard units silently. Say exactly which value is missing its unit and ask before running the calculation. If I mixed single-point and two-point data together, one rate constant and temperature alongside a second rate constant at a different temperature, tell me plainly that those are two different forms of the equation and ask which one I want solved, instead of guessing.
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Get Early AccessA rate constant problem built on the Arrhenius equation rarely goes wrong at the exponent. It goes wrong when a temperature stays in Celsius, or an activation energy answer gets reported in J/mol when the question wanted kJ/mol, or the other way around.
This tool solves k = A x e^(-Ea/RT) two ways. Give it your [KNOWN_VALUES], three of the four single-point quantities, rate constant, pre-exponential factor, activation energy, or temperature, and pick [SOLVE_FOR]. When the target sits inside the exponent, it takes the natural log of both sides first and isolates the variable before ever substituting a number. Or give it two rate constants measured at two different temperatures, and it solves for activation energy directly with the two-point form, no pre-exponential factor required.
Every temperature gets converted to Kelvin before it touches the equation, and every activation energy answer gets labeled clearly in both J/mol and kJ/mol so the two never get confused.
Ea is the energy hump on a reaction's energy diagram, the same hump that separates an exothermic or endothermic reaction from level ground. Once you can find Ea here, the ideal gas law solver covers the other half of what's happening to the molecules doing the colliding. Run it in the Dock Editor to keep the worked algebra next to your kinetics notes, or paste it into ChatGPT, Claude, or Gemini instead.
Run this in ChatGPT, Claude, Gemini, or the Dock Editor, then set [MODE] to single-point form when you have one data point with three known quantities, or two-point form when you have two rate constants measured at two temperatures.
In single-point form, set [SOLVE_FOR] to rate constant k, pre-exponential factor A, activation energy Ea, or temperature T. The tool takes the natural log of both sides first whenever Ea or T is the target, since both sit inside the exponent.
Drop your known numbers into [KNOWN_VALUES], with units attached. In two-point form that means k1, k2, T1, and T2. In single-point form it means whichever three of the four quantities you're not solving for.
Any temperature given in Celsius gets converted to Kelvin, adding 273.15, before it touches the equation. That conversion is shown as its own step.
An activation energy result is reported in J/mol first, then divided by 1000 and labeled again in kJ/mol, so you always know which unit you're reading.
Solve for activation energy from a single data point, or from two rate constants at two temperatures, with the log step and the Kelvin conversion both shown.
Practice the two-point form that shows up on free-response questions, where Ea has to come from two rate constants without ever knowing the pre-exponential factor.
Generate a model solution with the algebra isolation step kept visibly separate from the substitution step, useful for showing exactly where a student's answer diverged.
Drill unit discipline directly, Celsius versus Kelvin and J/mol versus kJ/mol, the two places an otherwise correct Arrhenius calculation most often loses points.
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