Enthalpy of Reaction (ΔrH) — Stoichiometric Calculations
The enthalpy of reaction, ΔrH, is the heat change at constant pressure
for the chemical equation as written. Its units are typically
kJ per mole of reaction (read: “kJ per balanced reaction”).
A negative value means the reaction releases heat (exothermic); a positive value means it
absorbs heat (endothermic).
Important: ΔrH refers to the balanced equation with its
smallest whole-number coefficients. If you multiply the entire equation by a factor
\(a\), the value of ΔrH also multiplies by \(a\). If you reverse the equation,
the sign of ΔrH changes.
How the math works
Suppose the balanced reaction is
\[
\nu_1\,\text{A}_1 + \nu_2\,\text{A}_2 + \cdots \;\longrightarrow\;
\nu_3\,\text{P}_1 + \nu_4\,\text{P}_2 + \cdots
\]
with stoichiometric coefficients \(\nu\) (all positive numbers). If a certain
“amount of reaction” \(f\) occurs (i.e., the balanced equation happens \(f\) times),
then:
\[
q = f\,\Delta_r H \qquad\text{and}\qquad
n_{\text{species}} = f\,\nu_{\text{species}} \;,
\]
where \(q\) is the heat (kJ) and \(n\) is the moles consumed (reactants) or formed (products).
We can determine \(f\) from either a species amount or from heat:
- From a species amount. If you know moles \(n_s\) of species \(s\) (or mass \(m_s\), using \(n_s=m_s/M_s\)),
then \(f = n_s/\nu_s\) and \(q = (n_s/\nu_s)\,\Delta_rH\).
- From heat. If you know \(q\), then \(f = q/\Delta_rH\) and the moles of species \(s\) are
\(n_s = |\nu_s|\,|q/\Delta_rH|\). Mass follows from \(m_s = n_s M_s\).
What the calculator expects
- Enter an unbalanced reaction. The tool balances it for you and shows the
normalized coefficients \(\nu\).
- Enter ΔrH in kJ per balanced reaction.
This is the same as “kJ per mole of reaction” used in many textbooks.
- Choose your task: (a) compute heat from a given amount of a chosen species (moles or grams),
or (b) compute the required amount of a chosen species from a specified heat \(q\) (J or kJ).
Worked example (matches the textbook idea)
Combustion of sucrose (balanced):
\[
\mathrm{C_{12}H_{22}O_{11}(s)} + 12\,\mathrm{O_2(g)} \;\longrightarrow\;
12\,\mathrm{CO_2(g)} + 11\,\mathrm{H_2O(l)}
\qquad \Delta_rH = -5.65\times 10^3\ \text{kJ (per reaction)}
\]
Question: How much heat is released by completely burning 1.00 kg of sucrose?
\[
n(\text{sucrose})=\frac{1000\ \text{g}}{342.3\ \text{g mol}^{-1}}=2.92\ \text{mol},
\qquad f=\frac{n}{\nu_{\text{sucrose}}}=\frac{2.92}{1}=2.92
\]
\[
q = f\,\Delta_rH = (2.92)\times(-5.65\times 10^3\ \text{kJ})
= -1.65\times 10^4\ \text{kJ}.
\]
The negative sign indicates heat is released. If you instead provide \(q\),
the calculator inverts these steps to return the moles/grams of any chosen species.
Units & sign conventions
- ΔrH units: kJ per mole of reaction (kJ per balanced reaction). The tool always reports
heat \(q\) in kJ. If you enter \(q\) in J, it will convert to kJ internally.
- Mass ↔ moles: \(n = m/M\) (with \(M\) in g·mol⁻¹). The tool uses standard atomic masses to compute \(M\).
- Exothermic vs. endothermic: ΔrH < 0 → exothermic (heat released); ΔrH > 0 → endothermic (heat absorbed).
Common pitfalls
- Wrong basis. Ensure your ΔrH matches the balanced equation shown by the tool. If a source
reports “kJ per mole of product X,” convert it to “per balanced reaction” by multiplying by the
coefficient of X in that equation.
- Scaled equations. If you double all coefficients, ΔrH must also be doubled.
- Sign mistakes. Keep the sign of ΔrH when converting between heat and amounts.
A negative \(q\) corresponds to heat released by the system.
Note on physical changes. For heating/cooling across phases, use
\(q = mc\Delta T\) for temperature changes and a phase enthalpy
(e.g., ΔHfus, ΔHvap) for plateaus. This calculator focuses on
reaction enthalpies supplied as ΔrH.
