Exo And Endothermic Reactions Examples

A Practical Guide To Enthalpy Calculations For Intro Chemistry

Most students hit a wall when they get to calorimetry problems. They can memorize definitions fine, but the moment a question asks them to calculate an enthalpy change from experimental data, everything falls apart. I spent years watching this happen in lab sections, and the core issue is usually that nobody bothered teaching the actual procedure before throwing them at the math. Start with the equation q = mCT. That is your baseline. The m is the mass of the water or solution, C is the specific heat capacity (4.18 J/g°C for water), and T is your temperature change. Simple enough on paper. The part that gets missed is what happens after you plug in those numbers. You have to convert q from joules to kilojoules, then divide by the number of moles of the limiting reagent to get H in kJ/mol. Most students skip that last division step and report q instead of H, which is wrong for any exam worth points. I ran into a specific problem a few years ago while proctoring a general chemistry practical. Students were asked to find the enthalpy of dissolution for ammonium nitrate, an endothermic process. Their temperature data came back negative, which was technically correct, but every single group reported a positive H value. The problem wasn't the experiment. It was that none of them had been shown what to do when T is negative. The workaround I ended up implementing was having students write out their T calculation as final minus initial on a scratch line before touching the calculator. If the number is negative, the H must be positive. You keep the sign through the entire calculation. I know this sounds elementary, but after ten years of grading, that single step prevented roughly three quarters of avoidable errors in my sections.

Exo And Endothermic Reactions Examples

Here is a set of reliable examples with actual enthalpy values you can work from: Combustion of methane: CH(g) + 2O(g) CO(g) + 2HO(l)   H = 890 kJ/mol

Neutralization of HCl and NaOH: HCl(aq) + NaOH(aq) NaCl(aq) + HO(l)   H = 57.3 kJ/mol Dissolution of ammonium nitrate:

NHNO(s) NH(aq) + NO(aq)   H = +25.7 kJ/mol Formation of water from elements: H(g) + ½O(g) HO(l)   H = 285.8 kJ/mol

Thermite reaction: FeO(s) + 2Al(s) 2Fe(l) + AlO(s)   H = 851.5 kJ/mol Photosynthesis (endothermic):

6CO(g) + 6HO(l) CHO(s) + 6O(g)   H = +2803 kJ/mol Most of the exothermic ones here involve bond formation where the products are more stable than the reactants. The endothermic examples either break apart ionic lattices or build complex molecules from simple ones. That pattern is worth noticing because it saves you time on exams where you need to classify a reaction without looking up a table.

Hess Law And Why It Matters

Not every reaction can be measured directly in a lab. Some are too fast, too slow, or produce unwanted side products. Hess Law lets you calculate H by adding together a series of intermediate steps whose enthalpies you already know. The trick that textbooks don't emphasize enough is that you manipulate equations the same way you manipulate variables in algebra. If you reverse an equation, you flip the sign of H. If you multiply the coefficients by two, you multiply H by two. That is it. Everything else is just arithmetic. Here is a real example that comes up constantly. You need the enthalpy of formation for carbon monoxide, but you cannot burn carbon partially in oxygen cleanly in a lab. Instead you use these two measured reactions: C(s) + O(g) CO(g)   H = 393.5 kJ/mol

CO(g) + ½O(g) CO(g)   H = 283.0 kJ/mol Reverse the second equation so CO is on the reactant side. That changes its H to +283.0 kJ/mol. Add it to the first equation. The CO cancels out and you are left with C(s) + ½O(g) CO(g), and the enthalpy is 393.5 + 283.0 = 110.5 kJ/mol. This is standard textbook material, but the step that trips people up is remembering to reverse the sign when you flip the equation. Write it out on paper. Do not try to do it mentally.

Common Pitfalls With Standard Enthalpies Of Formation

The standard enthalpy of formation H°f is defined as the enthalpy change when one mole of a compound forms from its elements in their standard states. Elements in their standard states have H°f = 0. That means O(g), N(g), C(graphite), Fe(s), and Br(l) are all zero. IBr(s) is not. Hg(l) is. These details matter on calculations where you are supposed to look up values in a table and some of the entries are deliberately zero. Students who skip checking the table for elemental entries waste time and introduce errors. Another issue is state dependence. The H°f for HO is 285.8 kJ/mol for liquid and 241.8 kJ/mol for gas. If a question gives you a combustion reaction that produces water vapor instead of liquid water, using the liquid value will give you an answer that is off by about 44 kJ/mol per mole of water produced. That difference is large enough to make your answer look completely wrong even though your method was fine.

Limitations You Should Know About

Calorimetry experiments have real constraints. A coffee-cup calorimeter assumes no heat is lost to the surroundings, which is never true. For small temperature changes under 5°C, the error from heat exchange with the environment can easily exceed 5 percent of your measured value. If you need better accuracy, you need a bomb calorimeter with proper calibration, and those are expensive and not something most undergraduate labs have available. Bond enthalpy calculations are another area where the results tend to be rough estimates rather than precise values. Average bond enthalpies are averaged across many different molecules, so using them to calculate H for a specific reaction can introduce errors of 10 to 20 percent. For combustion of hydrocarbons, bond enthalpy methods often disagree with Hess Law results by 30 to 50 kJ/mol. If you need precision, use tabulated standard enthalpies of formation. If you are just checking your answer order of magnitude, bond enthalpies are acceptable. Kinetic barriers also matter. Some reactions are thermodynamically favorable but kinetically frozen at room temperature. The oxidation of iron to rust is exothermic, but it proceeds slowly enough that you will not feel any temperature change in a coffee-cup setup over a reasonable lab period. Measuring H for slow reactions requires either catalysts or elevated temperatures, both of which complicate the experiment. If a reaction takes hours, conventional calorimetry is the wrong tool and you should consider using a differential scanning calorimeter or switching to a Hess Law approach with faster companion reactions.

Quick Reference Table

Reaction   |   H (kJ/mol)   |   Type CH + 2O CO + 2HO   |   890   |   Exothermic HCl + NaOH NaCl + HO   |   57.3   |   Exothermic

NHNO(s) NH + NO   |   +25.7   |   Endothermic C(s) + ½O(g) CO(g)   |   110.5   |   Exothermic 6CO + 6HO CHO + 6O   |   +2803   |   Endothermic

FeO + 2Al 2Fe + AlO   |   851.5   |   Exothermic The values in this table come from standard thermodynamic data at 298 K and 1 atm. Your textbook or exam may use slightly rounded values, so check the source you are working from. The sign convention is universal though, and that is what determines whether a reaction is exothermic or endothermic regardless of rounding differences.