Enthalpy Change Calculator: Stoichiometry & Thermochemistry

Precise calculation of reaction enthalpy using chemical equations and reactant amounts.

Calculate Reaction Enthalpy

What is Enthalpy Change Calculation Using Stoichiometry?

Calculating enthalpy change using stoichiometry is a fundamental concept in thermochemistry that bridges the macroscopic world of measurable quantities (like moles) with the microscopic world of chemical reactions and their associated energy transformations. It allows us to predict the heat absorbed or released during a chemical reaction when specific amounts of reactants are involved. This process is crucial for understanding reaction feasibility, designing chemical processes, and controlling energy output or input in industrial applications.

Who should use it: This calculation is essential for chemistry students learning about thermochemistry, chemical engineers designing reactors, research scientists studying reaction mechanisms, and anyone involved in processes where energy changes during chemical transformations are critical. It’s particularly useful when the standard enthalpy of reaction (ΔH°rxn) is known, but you need to determine the heat for a specific, non-standard quantity of reactants.

Common misconceptions:

• Enthalpy is always released (exothermic): Enthalpy change can be positive (endothermic, heat absorbed) or negative (exothermic, heat released). The sign is critical.
• Stoichiometry doesn’t affect energy: The amount of heat is directly proportional to the amount of substance reacting, as dictated by stoichiometry. Doubling the reactants (if they are limiting) doubles the heat.
• Any ΔH value can be used: The enthalpy change must be specific to the balanced chemical equation and correctly normalized (e.g., per mole of a specific reactant or product). Using an incorrect ΔH value will lead to inaccurate results.
• Reactant B is always used: If Reactant A is clearly the limiting reactant and the ΔH is given relative to Reactant A, calculations can proceed without explicit data for Reactant B, provided Reactant B is in excess or its coefficient is zero.

{primary_keyword} Formula and Mathematical Explanation

The core idea behind calculating enthalpy change using stoichiometry is that enthalpy is an extensive property, meaning it scales with the amount of substance. The standard enthalpy of reaction (ΔH°rxn) is typically given for the reaction as written with specific stoichiometric coefficients. To find the enthalpy change for a different quantity of reactants, we use the concept of a limiting reactant and scale the known ΔH°rxn accordingly.

Step-by-Step Derivation

1. Balanced Chemical Equation: Start with a correctly balanced chemical equation:
   
   `aA + bB → cC + dD`
   
   Where `a`, `b`, `c`, `d` are stoichiometric coefficients, and `A`, `B`, `C`, `D` are chemical species.
2. Standard Enthalpy of Reaction (ΔH°rxn): Obtain the standard enthalpy change for the reaction as written. This is often expressed in kJ per mole of reaction, meaning for every `a` moles of `A` reacting with `b` moles of `B` to form `c` moles of `C` and `d` moles of `D`. Sometimes, it’s given normalized to a specific reactant or product (e.g., kJ/mol of A).
3. Identify Limiting Reactant: Given the initial moles of reactants `moles_A` and `moles_B`, determine the limiting reactant.
   
   Calculate the molar ratio for each reactant:
   
   `Ratio_A = moles_A / a`
   
   `Ratio_B = moles_B / b` (if `b > 0`)
   
   The reactant with the smallest molar ratio is the limiting reactant. If `moles_B` or `b` is 0, Reactant A is assumed to be limiting or in excess relative to A.
4. Normalize Enthalpy Change: Ensure the enthalpy value used is consistent with the limiting reactant.
   • If `ΔH°rxn` is given for the reaction as written (per `a` moles of A and `b` moles of B):
   • If A is limiting: Use `ΔH_norm = ΔH°rxn / a` (kJ per mole of A reacted).
   • If B is limiting: Use `ΔH_norm = ΔH°rxn / b` (kJ per mole of B reacted).
   • If the input `enthalpyOfReactionPerMoleA` (let’s call it `ΔH_A`) is provided, it’s assumed to be the enthalpy change per mole of A reacting (i.e., `ΔH°rxn / a`).
   • If the input `enthalpyOfReactionPerMoleB` (let’s call it `ΔH_B`) is provided, it’s assumed to be the enthalpy change per mole of B reacting (i.e., `ΔH°rxn / b`).

   We will use the enthalpy value corresponding to the identified limiting reactant. If A is limiting, we use `ΔH_A`. If B is limiting, we prefer to use `ΔH_B` if provided and valid; otherwise, we might need to convert `ΔH_A` using the stoichiometric ratio (`ΔH_B = ΔH_A * (a / b)`). For simplicity, this calculator primarily relies on the enthalpy value associated with the limiting reactant’s input field.

5. Calculate Total Enthalpy Change: Multiply the moles of the limiting reactant by the normalized enthalpy change per mole of that limiting reactant.
   
   `ΔH_total = (Moles of Limiting Reactant) × ΔH_norm`
   
   If A is limiting: `ΔH_total = moles_A × ΔH_A`
   
   If B is limiting and `ΔH_B` is used: `ΔH_total = moles_B × ΔH_B`
   
   The unit will typically be kilojoules (kJ). A negative `ΔH_total` indicates an exothermic reaction (heat released), and a positive value indicates an endothermic reaction (heat absorbed).

Variable Explanations

Variable	Meaning	Unit	Typical Range
`a`, `b`, `c`, `d`	Stoichiometric coefficients in a balanced chemical equation	Unitless	Positive integers (often simple whole numbers)
`moles_A`, `moles_B`	Amount of substance of Reactant A and Reactant B available	moles (mol)	≥ 0
`Ratio_A`, `Ratio_B`	Molar ratio of available substance to stoichiometric coefficient	mol/mol (effectively unitless for comparison)	≥ 0
`ΔH°rxn`	Standard enthalpy change of reaction for the balanced equation as written	kJ/molrxn (kilojoules per mole of reaction)	Varies widely; can be positive or negative
`ΔH_A` or `enthalpyOfReactionPerMoleA`	Enthalpy change normalized per mole of Reactant A	kJ/mol	Varies widely; can be positive or negative
`ΔH_B` or `enthalpyOfReactionPerMoleB`	Enthalpy change normalized per mole of Reactant B	kJ/mol	Varies widely; can be positive or negative
`ΔH_total`	Total enthalpy change for the given amounts of reactants	kilojoules (kJ)	Varies widely; can be positive or negative

Practical Examples (Real-World Use Cases)

Example 1: Combustion of Methane

Consider the combustion of methane (CH₄):

`CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)`

The standard enthalpy change for this reaction is approximately -890.4 kJ per mole of CH₄ reacted.
Let’s assume we have 0.5 moles of CH₄ and an excess of O₂.

Inputs:

• Moles of Reactant A (CH₄): 0.5 mol
• Moles of Reactant B (O₂): (Excess, assumed sufficient)
• Stoichiometric Coefficient of Reactant A (CH₄): 1
• Stoichiometric Coefficient of Reactant B (O₂): 2
• Enthalpy Change per Mole of Reactant A (CH₄): -890.4 kJ/mol
• Enthalpy Change per Mole of Reactant B (O₂): 0 (or not needed as CH₄ is limiting)

Calculation:

• Reactant A (CH₄) is limiting because it’s the only one with a finite amount given, and it has excess O₂.
• Moles of Limiting Reactant = 0.5 mol CH₄
• Normalized Enthalpy = -890.4 kJ/mol CH₄
• Total Enthalpy Change (ΔH_total) = 0.5 mol × (-890.4 kJ/mol) = -445.2 kJ

Financial/Practical Interpretation:

This calculation shows that combusting 0.5 moles of methane will release 445.2 kJ of heat. This is crucial for calculating the energy output of natural gas burners, engines, or any process utilizing methane combustion. Understanding this energy release is vital for efficiency calculations and thermal management. This relates to potential energy efficiency metrics.

Example 2: Synthesis of Ammonia (Haber Process – Simplified)

Consider the synthesis of ammonia:

`N₂(g) + 3H₂(g) → 2NH₃(g)`

The standard enthalpy change (ΔH°rxn) is approximately -92.2 kJ per mole of N₂ reacted.
Suppose we have 10 moles of N₂ and 15 moles of H₂.

Inputs:

• Moles of Reactant A (N₂): 10 mol
• Moles of Reactant B (H₂): 15 mol
• Stoichiometric Coefficient of Reactant A (N₂): 1
• Stoichiometric Coefficient of Reactant B (H₂): 3
• Enthalpy Change per Mole of Reactant A (N₂): -92.2 kJ/mol
• Enthalpy Change per Mole of Reactant B (H₂): (Can be calculated as -92.2 / 3 ≈ -30.7 kJ/mol, but we’ll rely on N₂ being primary)

Calculation:

1. Determine Limiting Reactant:
   • Ratio_N₂ = 10 mol / 1 = 10 mol
   • Ratio_H₂ = 15 mol / 3 = 5 mol

   Since the ratio for H₂ (5 mol) is less than for N₂ (10 mol), H₂ is the limiting reactant.

2. Determine Normalized Enthalpy: We need the enthalpy change per mole of H₂. Since ΔH°rxn is -92.2 kJ per mole of N₂ (which corresponds to 3 moles of H₂), the enthalpy per mole of H₂ is -92.2 kJ / 3 mol H₂ ≈ -30.73 kJ/mol H₂. *Alternatively, if the calculator provides `enthalpyOfReactionPerMoleB` correctly, we would use that.* For this manual example, let’s use the calculated value.
3. Calculate Total Enthalpy Change:
   
   Moles of Limiting Reactant (H₂) = 15 mol
   
   Normalized Enthalpy (per mol H₂) ≈ -30.73 kJ/mol
   
   ΔH_total = 15 mol × (-30.73 kJ/mol) ≈ -460.95 kJ
   
   *(Note: Using the calculator’s logic where we prioritize the input for the limiting reactant:* If H₂ was designated Reactant A and its coeff 3, and input `enthalpyOfReactionPerMoleA` was -30.73, it would yield 15 * -30.73 = -460.95 kJ. If N₂ is A (coeff 1) and H₂ is B (coeff 3), and input `enthalpyOfReactionPerMoleA` is -92.2, and `enthalpyOfReactionPerMoleB` is 0 or not used, the limiting reactant calculation would correctly identify H₂ but might incorrectly use the N₂ enthalpy value if not carefully implemented. The calculator prioritizes the enthalpy associated with the identified limiting reactant based on its input field.* In this case, H₂ is limiting (ratio 5). The enthalpy per mole of H₂ is -92.2/3 = -30.7 kJ/mol. Total = 15 mol * -30.7 kJ/mol = -461 kJ. Let’s use the calculator’s logic: A=N2 (10 mol, coeff 1), B=H2 (15 mol, coeff 3). Ratio A = 10, Ratio B = 5. B is limiting. If `enthalpyOfReactionPerMoleA` is -92.2 and `enthalpyOfReactionPerMoleB` is 0, the calculator needs to calculate the correct enthalpy for B. A better approach is that the calculator uses the enthalpy value provided for the identified limiting reactant. If B is limiting, ideally `enthalpyOfReactionPerMoleB` should be -30.7. Since it’s 0, let’s recalculate assuming the user inputs `enthalpyOfReactionPerMoleA` as -92.2 for N2 and we need to find the enthalpy for H2. The calculator’s current logic prioritizes `enthalpyOfReactionPerMoleA` if A is limiting. If B is limiting, it checks `enthalpyOfReactionPerMoleB`. If that’s 0, it implies the user assumes it’s redundant or wants it calculated from A. Let’s assume the input `enthalpyOfReactionPerMoleA` (-92.2) is the primary value and the calculator will derive the enthalpy for the limiting reactant (H₂) from it.
   
   Correct logic: Limiting reactant is H₂ (15 mol). Enthalpy per mole of H₂ = ΔH°rxn / coeff(H₂) = -92.2 kJ / 3 = -30.73 kJ/mol. Total Enthalpy = 15 mol * -30.73 kJ/mol = -461 kJ.
   
   *Calculator’s simplified logic might use: Limiting Reactant Moles = 15 (from H2). Normalized Enthalpy = Use enthalpy associated with limiting reactant. If `enthalpyOfReactionPerMoleB` was -30.73, then Result = 15 * -30.73 = -461 kJ. If `enthalpyOfReactionPerMoleB` is 0, it might default to using `enthalpyOfReactionPerMoleA`, which is incorrect unless A is limiting.* For this example, let’s assume the user *intended* `enthalpyOfReactionPerMoleB` to be correctly calculated or provided. If we use the calculator structure strictly, and `enthalpyOfReactionPerMoleB` is 0, the `normalizedEnthalpy` would be 0, leading to 0 total enthalpy change, which is wrong. The calculator must handle this: if B is limiting and `enthalpyOfReactionPerMoleB` is 0, it should attempt to calculate it from A’s value if A’s value is valid. For now, let’s assume `enthalpyOfReactionPerMoleB` was correctly set to -30.73 for this example interpretation.)
   Let’s re-run assuming correct inputs for the example scenario:
   A=N₂ (10 mol, coeff 1), B=H₂ (15 mol, coeff 3).
   Ratio A = 10, Ratio B = 5. B is limiting.
   Input `enthalpyOfReactionPerMoleA` = -92.2 (for N₂)
   Input `enthalpyOfReactionPerMoleB` = -30.73 (for H₂)
   Limiting Reactant Moles = 15 mol (H₂)
   Normalized Enthalpy = -30.73 kJ/mol (from H₂)
   Total Enthalpy Change = 15 mol * -30.73 kJ/mol = -461 kJ.

Financial/Practical Interpretation:

The synthesis of ammonia is highly exothermic. Producing 15 moles of ammonia (which requires 15 moles of H₂ assuming N₂ is in excess) releases approximately 461 kJ of heat. This heat can be managed (recovered or dissipated) in industrial reactors. Understanding the energy balance is key to process optimization and safety in chemical process design.

How to Use This Enthalpy Change Calculator

Our Enthalpy Change Calculator simplifies the process of determining the heat absorbed or released in a chemical reaction based on the quantities of reactants involved. Follow these steps for accurate results:

1. Ensure Balanced Equation: Before using the calculator, make sure you have a correctly balanced chemical equation for the reaction you are studying. The stoichiometric coefficients are crucial.
2. Input Reactant Moles: Enter the available moles for Reactant A (`molesReactantA`) and, if applicable, Reactant B (`molesReactantB`). If only one reactant quantity is relevant or known, you can enter 0 for the other or ensure it’s in excess.
3. Enter Stoichiometric Coefficients: Input the corresponding stoichiometric coefficients for Reactant A (`stoichiometricCoefficientA`) and Reactant B (`stoichiometricCoefficientB`) exactly as they appear in the balanced equation. If a reactant isn’t involved or is in large excess (like a solvent), its coefficient might be 0 or irrelevant for limiting reactant calculation.
4. Provide Enthalpy Data:
   • Enter the enthalpy change of the reaction *per mole of Reactant A* (`enthalpyOfReactionPerMoleA`). This value should correspond to the enthalpy change when `stoichiometricCoefficientA` moles of A react. For example, if ΔH°rxn = -890.4 kJ for `1 CH₄ + 2 O₂ → …`, and A is CH₄, enter -890.4. If A were O₂, you’d need ΔH°rxn / 2.
   • If available, enter the enthalpy change *per mole of Reactant B* (`enthalpyOfReactionPerMoleB`). This is useful for verification or if B is the limiting reactant and the primary ΔH value was normalized to A.

   Important: Ensure the units are consistent (typically kJ/mol). Pay close attention to the sign: negative for exothermic (heat released), positive for endothermic (heat absorbed).

5. Calculate: Click the “Calculate Enthalpy Change” button.

How to Read Results:

• Primary Result (Total Enthalpy Change): This is the main output in kJ. A negative value means heat is released by the reaction; a positive value means heat is absorbed.
• Limiting Reactant Moles: Shows the calculated moles of the reactant that will be completely consumed first.
• Enthalpy Change per Mole of Limiting Reactant: This is the normalized enthalpy value used for the final calculation, specific to the limiting reactant.
• Stoichiometric Ratio: Displays the calculated molar ratios (available moles / coefficient) for each reactant, visually indicating which is limiting.
• Stoichiometry Table: Provides a clear breakdown of the limiting reactant calculation.
• Chart: Offers a visual representation of the relative contributions to enthalpy based on reactant quantities and normalized enthalpy values.
• Assumptions: Lists key assumptions made, such as standard conditions and the accuracy of input data.

Decision-Making Guidance:

The calculated `ΔH_total` is critical for:

• Energy Management: Determining if a reaction requires heating (endothermic) or generates heat that needs to be managed (exothermic).
• Process Scale-up: Estimating energy requirements or heat loads when scaling a reaction from lab to industrial quantities. This impacts equipment design for heating/cooling.
• Yield Prediction: While stoichiometry primarily dictates theoretical yield, understanding the energy balance can sometimes indirectly inform process efficiency and potential side reactions.
• Safety Analysis: Highly exothermic reactions pose risks if heat is not dissipated effectively.

Understanding these calculations is fundamental for efficient and safe chemical reaction engineering.

Key Factors That Affect Enthalpy Change Results

Several factors can influence the accuracy and applicability of calculated enthalpy changes:

1. Accuracy of Input Data: The most significant factor. Errors in measured moles, stoichiometric coefficients, or the provided standard enthalpy values (`ΔH°rxn`, `ΔH_A`, `ΔH_B`) will directly lead to incorrect results. Precise measurements are paramount.
2. Balanced Chemical Equation: An incorrect or unbalanced chemical equation will result in wrong stoichiometric coefficients and an inaccurate limiting reactant determination, rendering the enthalpy calculation invalid. Ensure the equation follows the law of conservation of mass.
3. Limiting Reactant Identification: Correctly identifying the limiting reactant is essential. If the wrong reactant is assumed limiting, the calculated `ΔH_total` will be incorrect, as the extent of reaction is determined by the limiting species.
4. Normalization of Enthalpy Change: The enthalpy value used must be correctly normalized. If `ΔH°rxn` is given per mole of reaction, it must be divided by the coefficient of the limiting reactant. If the input `enthalpyOfReactionPerMoleA` is incorrectly provided (e.g., not truly per mole of A, or a typo), the calculation will be flawed. This is critical for accurate thermochemical calculations.
5. Temperature and Pressure (Standard vs. Actual Conditions): Standard enthalpy values (`ΔH°`) are typically measured under specific standard conditions (e.g., 298 K and 1 atm or 1 bar). Real-world reactions may occur under different T/P, and enthalpy changes can vary with these conditions (though often modestly for many reactions unless phase changes or significant temperature differences are involved). Hess’s Law can help adjust for temperature differences, but this calculator assumes standard enthalpy values apply.
6. Phase Changes: The provided enthalpy values must account for the physical states (solid, liquid, gas) of reactants and products. If a reaction involves a phase change (e.g., water boiling), the enthalpy of that change must be included in the overall ΔH°rxn. This calculator assumes the provided enthalpy data is comprehensive for the specified states.
7. Side Reactions and Impurities: Real chemical processes often involve unintended side reactions or impurities in reactants. These can consume reactants, affect the overall energy balance, and produce different products, deviating from the idealized calculation based on a single, clean reaction.
8. Heat Loss/Gain to Surroundings: This calculation determines the theoretical enthalpy change *of the reaction itself*. In practice, some heat may be lost to or gained from the surroundings, affecting the measured temperature change of the system. This calculator focuses solely on the reaction’s intrinsic enthalpy change. Effective insulation and heat exchange are key in industrial process control.

Frequently Asked Questions (FAQ)

What is the difference between enthalpy change and heat?

Enthalpy change (ΔH) is a thermodynamic state function representing the total heat content change of a system at constant pressure. Heat (q) is the transfer of thermal energy between systems. Under constant pressure conditions, the heat exchanged (q_p) is equal to the enthalpy change (ΔH). So, while often used interchangeably in this context, ΔH specifically refers to the heat change associated with a process occurring at constant pressure.

Can I use mass instead of moles?

No, enthalpy changes are fundamentally related to the number of moles reacting, not mass directly. You must first convert the mass of your reactants to moles using their respective molar masses before using this calculator. Mass is proportional to moles, but the proportionality constant (molar mass) differs for each substance.

What does a negative enthalpy change signify?

A negative enthalpy change (ΔH < 0) signifies an exothermic reaction. This means the reaction releases energy, usually in the form of heat, into the surroundings. The products have lower enthalpy than the reactants.

What does a positive enthalpy change signify?

A positive enthalpy change (ΔH > 0) signifies an endothermic reaction. This means the reaction absorbs energy, usually in the form of heat, from the surroundings. The products have higher enthalpy than the reactants.

How do I find the standard enthalpy of reaction (ΔH°rxn)?

Standard enthalpies of reaction can be found in chemical data tables, textbooks, online databases (like NIST WebBook), or calculated using standard enthalpies of formation (ΔH°f) via Hess’s Law: ΔH°rxn = Σ(ΔH°f of products) – Σ(ΔH°f of reactants). Ensure the values correspond to the correct states (gaseous, liquid, solid, aqueous).

What if Reactant B is not involved in the reaction?

If Reactant B is not part of the stoichiometric reaction (e.g., it’s a catalyst, solvent, or simply not a reactant), set its `Stoichiometric Coefficient` to 0 and its `Moles` to 0 or a value indicating excess. The calculator will correctly identify Reactant A as the basis for calculation if its coefficient is non-zero.

Does this calculator account for reaction kinetics?

No, this calculator deals strictly with thermochemistry (enthalpy changes), which is governed by the initial and final states of the reaction, not the speed (kinetics) at which it occurs. An endothermic reaction might be very slow even if it has a large positive ΔH.

How accurate are the results?

The accuracy of the results is directly dependent on the accuracy of the input values: the moles of reactants, the stoichiometric coefficients from the balanced equation, and crucially, the provided standard enthalpy change value. The calculator performs the mathematical conversion accurately, but “garbage in, garbage out” applies. For precise industrial applications, experimentally determined enthalpy values under operating conditions might be necessary.

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// Since external libraries are forbidden per instructions, we must assume Chart.js is available or implement drawing manually.
// Given the strict constraint "NO external chart libraries", I will REMOVE Chart.js and implement a basic SVG or Canvas drawing if possible.
// Re-evaluation: The prompt says "NO external chart libraries" but then lists native

// --- REPLACING CHART.JS WITH NATIVE CANVAS DRAWING ---
// The 'drawChart' function above needs to be re-written to use canvas context API directly.
// This is significantly more complex than using Chart.js.
// Let's simplify the chart requirement for native canvas.

function drawChart(molesA, molesB, coeffA, coeffB, enthalpyA, enthalpyB, limitingMoles, normalizedEnthalpy) {
var canvas = document.getElementById('enthalpyChart');
var ctx = canvas.getContext('2d');
canvas.width = canvas.clientWidth; // Adjust canvas size to its container
canvas.height = 300; // Fixed height for consistency, or make responsive

ctx.clearRect(0, 0, canvas.width, canvas.height); // Clear previous drawing

var chartWidth = canvas.width;
var chartHeight = canvas.height;
var barPadding = 10;
var categoryPadding = 20;
var legendHeight = 40; // Space for legend
var titleHeight = 30; // Space for title

var usableWidth = chartWidth - 2 * barPadding;
var usableHeight = chartHeight - titleHeight - legendHeight - 2 * barPadding;

var labels = ['Reactant A', 'Reactant B', 'Product Formation (Illustrative)'];
var data = [];
var colors = ['#ff6347', '#4682b4', '#28a745']; // Tomato, SteelBlue, Success Green

var totalCalculatedEnthalpy = parseFloat(document.getElementById('primaryResultValue').textContent);

var valueA = 0;
var valueB = 0;
var valueC = 0; // Product formation

// Determine which reactant is limiting and calculate its contribution
var isLimitingA = false;
var isLimitingB = false;
var ratioA = coeffA > 0 ? molesA / coeffA : Infinity;
var ratioB = coeffB > 0 ? molesB / coeffB : Infinity;

if (!isNaN(ratioA) && !isNaN(ratioB)) {
if (ratioA <= ratioB) { // A is limiting or B is excess/NA isLimitingA = true; valueA = normalizedEnthalpy !== 0 ? normalizedEnthalpy : (enthalpyA !== 0 ? enthalpyA : 0); // Use normalized or enthalpyA if (valueA === 0 && enthalpyA !== 0 && coeffA > 0) { // Attempt calculation if needed
valueA = enthalpyA; // Use enthalpyA directly if normalizedEnthalpy is 0
}
valueA = valueA * molesA; // Final kJ value for A
} else { // B is limiting
isLimitingB = true;
var enthalpyPerMoleB = 0;
if (enthalpyB !== 0) {
enthalpyPerMoleB = enthalpyB;
} else if (enthalpyA !== 0 && coeffA > 0 && coeffB > 0) {
enthalpyPerMoleB = enthalpyA * (coeffA / coeffB);
}
valueB = enthalpyPerMoleB * molesB;
}
} else if (!isNaN(ratioA) && coeffA > 0) { // Only A has valid ratio
isLimitingA = true;
valueA = normalizedEnthalpy !== 0 ? normalizedEnthalpy : (enthalpyA !== 0 ? enthalpyA : 0);
if (valueA === 0 && enthalpyA !== 0 && coeffA > 0) {
valueA = enthalpyA;
}
valueA = valueA * molesA;
}

// Illustrative product contribution
if (!isNaN(totalCalculatedEnthalpy) && totalCalculatedEnthalpy !== 0) {
valueC = -totalCalculatedEnthalpy * 0.6; // Make it a significant illustrative value
}

data.push(valueA, valueB, valueC);

// Find min and max for scaling the y-axis
var minValue = Math.min(0, ...data);
var maxValue = Math.max(0, ...data);
var range = maxValue - minValue;

if (range === 0) range = 1; // Avoid division by zero if all values are 0

// Draw Title
ctx.fillStyle = '#004a99';
ctx.font = 'bold 16px Segoe UI';
ctx.textAlign = 'center';
ctx.fillText('Conceptual Enthalpy Contributions', chartWidth / 2, barPadding + 15);

// Draw Bars
var numBars = labels.length;
var totalBarWidth = usableWidth - (numBars - 1) * categoryPadding;
var barWidth = totalBarWidth / numBars;

for (var i = 0; i < numBars; i++) { var barHeight = Math.abs(data[i] / range) * (usableHeight * 0.8); // Use 80% of usable height for bars var x = barPadding + i * (barWidth + categoryPadding); var y = chartHeight - legendHeight - barPadding - (data[i] >= 0 ? barHeight : 0); // Base for positive bars

// Adjust y for negative bars (below zero line)
if (data[i] < 0) { y = chartHeight - legendHeight - barPadding - barHeight; barHeight = Math.abs(barHeight); // Ensure height is positive } else { y = chartHeight - legendHeight - barPadding - barHeight; // Position from bottom up } // Center bars within their category space x = barPadding + i * (barWidth + categoryPadding) + (categoryPadding / 2); // Draw the bar ctx.fillStyle = colors[i]; ctx.fillRect(x, y, barWidth, barHeight); // Draw label below bar ctx.fillStyle = '#333'; ctx.font = '12px Segoe UI'; ctx.textAlign = 'center'; ctx.fillText(labels[i], x + barWidth / 2, chartHeight - legendHeight + 15); // Draw value above bar ctx.textAlign = 'center'; ctx.fillText(data[i].toFixed(2) + ' kJ', x + barWidth / 2, y - 5); } // Draw Y-axis labels (simplified) ctx.fillStyle = '#333'; ctx.font = '12px Segoe UI'; ctx.textAlign = 'right'; var zeroLineY = chartHeight - legendHeight - barPadding - (0 - minValue) / range * (usableHeight * 0.8); ctx.fillText(maxValue.toFixed(2) + ' kJ', barPadding - 5, titleHeight + barPadding); ctx.fillText('0 kJ', barPadding - 5, zeroLineY); ctx.fillText(minValue.toFixed(2) + ' kJ', barPadding - 5, chartHeight - legendHeight - barPadding); // Draw Y-axis line ctx.beginPath(); ctx.moveTo(barPadding, titleHeight); ctx.lineTo(barPadding, chartHeight - legendHeight - barPadding); ctx.strokeStyle = '#ccc'; ctx.stroke(); // Draw legend (manually) var legendX = chartWidth / 2; var legendY = chartHeight - legendHeight + 10; ctx.textAlign = 'left'; ctx.font = '12px Segoe UI'; // Reactant A legend ctx.fillStyle = colors[0]; ctx.fillRect(legendX - 80, legendY, 12, 12); ctx.fillStyle = '#333'; ctx.fillText('Reactant Enthalpy Contribution', legendX - 60, legendY + 10); // Reactant B legend ctx.fillStyle = colors[1]; ctx.fillRect(legendX - 80, legendY + 20, 12, 12); ctx.fillStyle = '#333'; ctx.fillText('Product Enthalpy Contribution (Illustrative)', legendX - 60, legendY + 30); // Make sure the chart is updated when window resizes (basic implementation) window.addEventListener('resize', function() { // Re-run drawChart on resize - might need debouncing for performance setTimeout(function() { calculateEnthalpy(); // Recalculate to get updated values and trigger redraw }, 100); }); }