Next, we see that F2 is also needed as a reactant. Watch Video \(\PageIndex{1}\) to see these steps put into action while solving example \(\PageIndex{1}\). We saw in the balanced equation that one mole of ethanol reacts with three moles of oxygen gas. Since summing these three modified reactions yields the reaction of interest, summing the three modified H values will give the desired H: Aluminum chloride can be formed from its elements: (i) \(\ce{2Al}(s)+\ce{3Cl2}(g)\ce{2AlCl3}(s)\hspace{20px}H=\:?\), (ii) \(\ce{HCl}(g)\ce{HCl}(aq)\hspace{20px}H^\circ_{(ii)}=\mathrm{74.8\:kJ}\), (iii) \(\ce{H2}(g)+\ce{Cl2}(g)\ce{2HCl}(g)\hspace{20px}H^\circ_{(iii)}=\mathrm{185\:kJ}\), (iv) \(\ce{AlCl3}(aq)\ce{AlCl3}(s)\hspace{20px}H^\circ_{(iv)}=\mathrm{+323\:kJ/mol}\), (v) \(\ce{2Al}(s)+\ce{6HCl}(aq)\ce{2AlCl3}(aq)+\ce{3H2}(g)\hspace{20px}H^\circ_{(v)}=\mathrm{1049\:kJ}\). If methanol is burned in air, we have: \[\ce{CH_3OH} + \ce{O_2} \rightarrow \ce{CO_2} + 2 \ce{H_2O} \: \: \: \: \: He = 890 \: \text{kJ/mol}\nonumber \]. Now, when we multiply through the moles of carbon-carbon single bonds, cancel and this gives us From data tables find equations that have all the reactants and products in them for which you have enthalpies. We also can use Hesss law to determine the enthalpy change of any reaction if the corresponding enthalpies of formation of the reactants and products are available. You will need to understand why it works..Hess Law states that the enthalpies of the products and the reactants are the same, All tip submissions are carefully reviewed before being published. It shows how we can find many standard enthalpies of formation (and other values of H) if they are difficult to determine experimentally. So let's write in here, the bond enthalpy for Figure \(\PageIndex{2}\): The steps of example \(\PageIndex{1}\) expressed as an energy cycle. closely to dots structures or just look closely This book uses the The burning of ethanol produces a significant amount of heat. (a) Write the balanced equation for the combustion of ethanol to CO 2 (g) and H 2 O(g), and, using the data in Appendix G, calculate the enthalpy of combustion of 1 mole of ethanol. \[\begin{align} \text{equation 1: } \; \; \; \; & P_4+5O_2 \rightarrow \textcolor{red}{2P_2O_5} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \;\; \; \; \;\Delta H_1 \nonumber \\ \text{equation 2: } \; \; \; \; & \textcolor{red}{2P_2O_5} +6H_2O \rightarrow 4H_3PO_4 \; \; \; \; \; \; \; \; \Delta H_2 \nonumber\\ \nonumber \\ \text{equation 3: } \; \; \; \; & P_4 +5O_2 + 6H_2O \rightarrow 3H_3PO_4 \; \; \; \; \Delta H_3 \end{align}\]. Except where otherwise noted, textbooks on this site We will consider how to determine the amount of work involved in a chemical or physical change in the chapter on thermodynamics. to sum the bond enthalpies of the bonds that are formed. Table \(\PageIndex{1}\) Heats of combustion for some common substances. In the above equation the P2O5 is an intermediate, and if we add the two equations the intermediate can cancel out. Be sure to take both stoichiometry and limiting reactants into account when determining the H for a chemical reaction. Convert into kJ by dividing q by 1000. A type of work called expansion work (or pressure-volume work) occurs when a system pushes back the surroundings against a restraining pressure, or when the surroundings compress the system. We recommend using a The heat of combustion refers to the energy that is released as heat when a compound undergoes complete combustion with oxygen under standard conditions. Assume that coffee has the same specific heat as water. In section 5.6.3 we learned about bomb calorimetry and enthalpies of combustion, and table \(\PageIndex{1}\) contains some molar enthalpy of combustion data. So, identify species that only exist in one of the given equations and put them on the desired side of the equation you want to produce, following the Tips above. So to represent those two moles, I've drawn in here, two molecules of CO2. Calculate the enthalpy of combustion of exactly 1 L of ethanol. One of the values of enthalpies of formation is that we can use them and Hess's Law to calculate the enthalpy change for a reaction that is difficult to measure, or even dangerous. You calculate #H_"c"^# from standard enthalpies of formation: #H_"c"^o = H_"f"^"(p)" - H_"f"^"(r)"#. This can be obtained by multiplying reaction (iii) by \(\frac{1}{2}\), which means that the H change is also multiplied by \(\frac{1}{2}\): \[\ce{ClF}(g)+\frac{1}{2}\ce{O2}(g)\frac{1}{2}\ce{Cl2O}(g)+\frac{1}{2}\ce{OF2}(g)\hspace{20px} H=\frac{1}{2}(205.6)=+102.8\: \ce{kJ} \nonumber\]. A blank line = 1 or you can put in the 1 that is fine. For example, when 1 mole of hydrogen gas and 1212 mole of oxygen gas change to 1 mole of liquid water at the same temperature and pressure, 286 kJ of heat are released. This H value indicates the amount of heat associated with the reaction involving the number of moles of reactants and products as shown in the chemical equation. Hess's Law When we add these together, we get 5,974. Subtract the reactant sum from the product sum. If you stand on the summit of Mt. Because enthalpy of reaction is a state function the energy change between reactants and products is independent of the path. the!heat!as!well.!! Table \(\PageIndex{2}\): Standard enthalpies of formation for select substances. 3 Put the substance at the base of the standing rod. % of people told us that this article helped them. 447 kJ B. The following conventions apply when using H: A negative value of an enthalpy change, H < 0, indicates an exothermic reaction; a positive value, H > 0, indicates an endothermic reaction. The following tips should make these calculations easier to perform. (a) Assuming that coke has the same enthalpy of formation as graphite, calculate \({\bf{\Delta H}}_{{\bf{298}}}^{\bf{0}}\)for this reaction. Since summing these three modified reactions yields the reaction of interest, summing the three modified H values will give the desired H: (i) 2Al(s)+3Cl2(g)2AlCl3(s)H=?2Al(s)+3Cl2(g)2AlCl3(s)H=? The molar heat of combustion corresponds to the energy released, in the form of heat, in a combustion reaction of 1 mole of a substance. structures were broken and all of the bonds that we drew in the dot For example, given that: Then, for the reverse reaction, the enthalpy change is also reversed: Looking at the reactions, we see that the reaction for which we want to find H is the sum of the two reactions with known H values, so we must sum their Hs: The enthalpy of formation, Hf,Hf, of FeCl3(s) is 399.5 kJ/mol. Enthalpies of formation are usually found in a table from CRC Handbook of Chemistry and Physics. Note, if two tables give substantially different values, you need to check the standard states. (b) Methanol, a liquid fuel that could possibly replace gasoline, can be prepared from water gas and additional hydrogen at high temperature and pressure in the presence of a suitable catalyst:\({\bf{2}}{{\bf{H}}_{\bf{2}}}\left( {\bf{g}} \right){\bf{ + CO}}\left( {\bf{g}} \right) \to {\bf{C}}{{\bf{H}}_{\bf{3}}}{\bf{OH}}\left( {\bf{g}} \right)\). So for the combustion of one mole of ethanol, 1,255 kilojoules of energy are released. (The symbol H is used to indicate an enthalpy change for a reaction occurring under nonstandard conditions. Assume that the coffee has the same density and specific heat as water. The result is shown in Figure 5.24. Research source. 348 kilojoules per mole of reaction. times the bond enthalpy of an oxygen-oxygen double bond. &\ce{ClF}(g)+\frac{1}{2}\ce{O2}(g)\frac{1}{2}\ce{Cl2O}(g)+\frac{1}{2}\ce{OF2}(g)&&H=\mathrm{+102.8\: kJ}\\ Do not include units in you answer C2H2 (g) + O2 (g) - 2C02 (g) + H20 (9) Bond C-C CEC Bond Energy (kJ/mol) 347 614 839 C-H C=0 O-H This problem has been solved! carbon-oxygen double bonds. Next, we do the same thing for the bond enthalpies of the bonds that are formed. The heat of combustion of acetylene is -1309.5 kJ/mol. And that's about 413 kilojoules per mole of carbon-hydrogen bonds. Its energy contentis H o combustion = -1212.8kcal/mole. Direct link to Morteza Aslami's post what do we mean by bond e, Posted a month ago. So to this, we're going to write in here, a five, and then the bond enthalpy of a carbon-hydrogen bond. , Calculate the grams of O2 required for the combustion of 25.9 g of ethylcyclopentane, A 32.0 L cylinder containing helium gas at a pressure of 38.5 atm is used to fill a weather balloon in order to lift equipment into the stratosphere. It shows how we can find many standard enthalpies of formation (and other values of H) if they are difficult to determine experimentally. Enthalpy values for specific substances cannot be measured directly; only enthalpy changes for chemical or physical processes can be determined. 3: } \; \; \; \; & C_2H_6+ 3/2O_2 \rightarrow 2CO_2 + 3H_2O \; \; \; \; \; \Delta H_3= -1560 kJ/mol \end{align}\], Video \(\PageIndex{1}\) shows how to tackle this problem. We did this problem, assuming that all of the bonds that we drew in our dots This equation says that 85.8 kJ is of energy is exothermically released when one mole of liquid water is formed by reacting one mole of hydrogen gas and 1/2mol oxygen gas (3.011x1023 molecules of O2). If we scrutinise this statement: "the total energies of the products being less than the reactants", then a negative enthalpy cannot be an exothermic. and then the product of that reaction in turn reacts with water to form phosphorus acid. The provided amounts of the two reactants are, The provided molar ratio of perchlorate-to-sucrose is then. We can choose a hypothetical two step path where the atoms in the reactants are broken into the standard state of their element (left side of Figure \(\PageIndex{3}\)), and then from this hypothetical state recombine to form the products (right side of Figure \(\PageIndex{3}\)). To figure out which bonds are broken and which bonds are formed, it's helpful to look at the dot structures for our molecules. structures were formed. negative sign in here because this energy is given off. The bonds enthalpy for an oxygen hydrogen single bond is 463 kilojoules per mole, and we multiply that by six. However, if we look What is important here, is that by measuring the heats of combustion scientists could acquire data that could then be used to predict the enthalpy of a reaction that they may not be able to directly measure. Because the H of a reaction changes very little with such small changes in pressure (1 bar = 0.987 atm), H values (except for the most precisely measured values) are essentially the same under both sets of standard conditions. in the gaseous state. Conversely, energy is transferred out of a system when heat is lost from the system, or when the system does work on the surroundings. Find the amount of substance burned by subtracting the final mass from the initial mass of the substance in g. Divide q in kJ by the mass of the substance burned. So let's start with the ethanol molecule. It has a high octane rating and burns more slowly than regular gas. H for a reaction in one direction is equal in magnitude and opposite in sign to H for the reaction in the reverse direction. Note the enthalpy of formation is a molar function, so you can have non-integer coefficients. [1] The Heat of Combustion of a substance is defined as the amount of energy in the form of heat is liberated when an amount of the substance undergoes combustion. In this section we will use Hess's law to use combustion data to calculate the enthalpy of reaction for a reaction we never measured. So to this, we're going to add six The heat (enthalpy) of combustion of acetylene = -1228 kJ The heat of combustion refers to the amount of heat released when 1 mole of a substance is burned. However, we often find it more useful to divide one extensive property (H) by another (amount of substance), and report a per-amount intensive value of H, often normalized to a per-mole basis. And we can see in each molecule of O2, there's an oxygen-oxygen double bond. For the formation of 2 mol of O3(g), H=+286 kJ.H=+286 kJ. \[\ce{N2}(g)+\ce{2O2}(g)\ce{2NO2}(g) \nonumber\], \[\ce{N2}(g)+\ce{O2}(g)\ce{2NO}(g)\hspace{20px}H=\mathrm{180.5\:kJ} \nonumber\], \[\ce{NO}(g)+\frac{1}{2}\ce{O2}(g)\ce{NO2}(g)\hspace{20px}H=\mathrm{57.06\:kJ} \nonumber\]. Summing these reaction equations gives the reaction we are interested in: Summing their enthalpy changes gives the value we want to determine: So the standard enthalpy change for this reaction is H = 138.4 kJ. When you multiply these two together, the moles of carbon-carbon When we do this, we get positive 4,719 kilojoules. Note, these are negative because combustion is an exothermic reaction. So that's a total of four See Answer Calculate the molar enthalpy of formation from combustion data using Hess's Law Using the enthalpy of formation, calculate the unknown enthalpy of the overall reaction Calculate the heat evolved/absorbed given the masses (or volumes) of reactants. (b) The density of ethanol is 0.7893 g/mL. wikiHow is a wiki, similar to Wikipedia, which means that many of our articles are co-written by multiple authors. Let's use bond enthalpies to estimate the enthalpy of combustion of ethanol. In this case, there is no water and no carbon dioxide formed. Many readily available substances with large enthalpies of combustion are used as fuels, including hydrogen, carbon (as coal or charcoal), and hydrocarbons (compounds containing only hydrogen and carbon), such as methane, propane, and the major components of gasoline. (This amount of energy is enough to melt 99.2 kg, or about 218 lbs, of ice.) Free and expert-verified textbook solutions. By definition, the standard enthalpy of formation of an element in its most stable form is equal to zero under standard conditions, which is 1 atm for gases and 1 M for solutions. If 1 mol of acetylene produces -1301.1 kJ, then 4.8 mol of acetylene produces: \(\begin{array}{l}{\rm{ = 1301}}{\rm{.1 \times 4}}{\rm{.8 }}\\{\rm{ = 6245}}{\rm{.28 kJ }}\\{\rm{ = 6}}{\rm{.25 kJ}}\end{array}\). . The one is referring to breaking one mole of carbon-carbon single bonds. In fact, it is not even a combustion reaction. Question: Calculate the heat capacity, in joules and in calories per degree, of the following: The following sequence of reactions occurs in the commercial production of aqueous nitric acid: 4NH3(g) + 5O2(g) 4NO(g) + 6H2O(l) H = 907 kJ, 3NO2 + H2O(l) 2HNO3(aq) + NO(g) H = 139 kJ. Right now, we're summing an endothermic reaction. carbon-oxygen single bond. In efforts to reduce gas consumption from oil, ethanol is often added to regular gasoline. So we have one carbon-carbon bond. -1228 kJ C. This problem has been solved! The heat of combustion refers to the amount of heat released when 1 mole of a substance is burned. Estimate the heat of combustion for one mole of acetylene: C2H2 (g) + O2 (g) 2CO2 (g) + H2O (g) Bond Bond Energy/ (kJ/mol CC 839 C-H 413 O=O 495 C=O 799 O-H 467 A. At this temperature, Hvalues for CO2(g) and H2O(l) are -393 and -286 kJ/mol, respectively. #DeltaH_("C"_2"H"_2"(g)")^o = "226.73 kJ/mol"#; #DeltaH_("CO"_2"(g)")^o = "-393.5 kJ/mol"#; #DeltaH_("H"_2"O(l)")^o = "-285.8 kJ/mol"#, #"[2 (-393.5) + (-295.8)] [226.7 + 0] kJ" = "-1082.8 - 226.7" =#. Measure the temperature of the water and note it in degrees celsius. The molar heat of combustion \(\left( He \right)\) is the heat released when one mole of a substance is completely burned. around the world. while above we got -136, noting these are correct to the first insignificant digit. Hcomb (H2(g)) = -276kJ/mol, Note, in the following video we used Hess's Law to calculate the enthalpy for the balanced equation, with integer coefficients. The enthalpy of formation, \(H^\circ_\ce{f}\), of FeCl3(s) is 399.5 kJ/mol. The reaction of gasoline and oxygen is exothermic. The enthalpy change for this reaction is 5960 kJ, and the thermochemical equation is: Enthalpy changes are typically tabulated for reactions in which both the reactants and products are at the same conditions. H -84 -(52.4) -0= -136.4 kJ. The calculator estimates the cost for each fuel type to deliver 100,000 BTU's of heat to your house. five times the bond enthalpy of an oxygen-hydrogen single bond. The heat(enthalpy) of combustion of acetylene = 2902.5 kJ - 4130 kJ, The heat(enthalpy) of combustion of acetylene = -1227.5 kJ. After 5 minutes, both the metal and the water have reached the same temperature: 29.7 C. Creative Commons Attribution/Non-Commercial/Share-Alike. So this was 348 kilojoules per one mole of carbon-carbon single bonds. The reaction of acetylene with oxygen is as follows: C 2 H 2 ( g) + 5 2 O 2 ( g) 2 C O 2 ( g) + H 2 O ( l) Here, in the above reaction, one mole of acetylene produces -1301.1 kJ heat. The calculator takes into account the cost of the fuel, energy content of the fuel, and the efficiency of your furnace. \[\Delta H_{reaction}=\sum m_i \Delta H_{f}^{o}(products) - \sum n_i \Delta H_{f}^{o}(reactants) \nonumber \]. Note that this result was obtained by (1) multiplying the HfHf of each product by its stoichiometric coefficient and summing those values, (2) multiplying the HfHf of each reactant by its stoichiometric coefficient and summing those values, and then (3) subtracting the result found in (2) from the result found in (1). The reaction of acetylene with oxygen is as follows: \({{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{2}}}{\rm{(g) + }}\frac{{\rm{5}}}{{\rm{2}}}{{\rm{O}}_{\rm{2}}}{\rm{(g)}} \to {\rm{2C}}{{\rm{O}}_{\rm{2}}}{\rm{(g) + }}{{\rm{H}}_{\rm{2}}}{\rm{O(l)}}\). What are the units used for the ideal gas law? Coupled Equations: A balanced chemical equation usually does not describe how a reaction occurs, that is, its mechanism, but simply the number of reactants in products that are required for mass to be conserved. Do the same for the reactants. The distance you traveled to the top of Kilimanjaro, however, is not a state function. And we're multiplying this by five. Want to cite, share, or modify this book? Calculate Hfor acetylene. Water gas, a mixture of \({{\bf{H}}_{\bf{2}}}\) and CO, is an important industrial fuel produced by the reaction of steam with red hot coke, essentially pure carbon:\({\bf{C}}\left( {\bf{s}} \right){\bf{ + }}{{\bf{H}}_{\bf{2}}}{\bf{O}}\left( {\bf{g}} \right) \to {\bf{CO}}\left( {\bf{g}} \right){\bf{ + }}{{\bf{H}}_{\bf{2}}}\left( {\bf{g}} \right)\). Both have the same change in elevation (altitude or elevation on a mountain is a state function; it does not depend on path), but they have very different distances traveled (distance walked is not a state function; it depends on the path). So for the final standard change in enthalpy for our chemical reaction, it's positive 4,719 minus 5,974, which gives us negative 1,255 kilojoules. Write the equation you want on the top of your paper, and draw a line under it. Substances act as reservoirs of energy, meaning that energy can be added to them or removed from them. This calculator provides a quick way to compare the cost and CO2 emissions for various fuels. Calculations using the molar heat of combustion are described. On the other hand, the heat produced by a reaction measured in a bomb calorimeter (Figure 5.17) is not equal to H because the closed, constant-volume metal container prevents the pressure from remaining constant (it may increase or decrease if the reaction yields increased or decreased amounts of gaseous species). Robert E. Belford (University of Arkansas Little Rock; Department of Chemistry). That is, the energy lost in the exothermic steps of the cycle must be regained in the endothermic steps, no matter what those steps are. how much heat is produced by the combustion of 125 g of acetylene c2h2. a one as the coefficient in front of ethanol. By the end of this section, you will be able to: Thermochemistry is a branch of chemical thermodynamics, the science that deals with the relationships between heat, work, and other forms of energy in the context of chemical and physical processes. The heat(enthalpy) of combustion of acetylene = -1228 kJ. Explain how you can confidently determine the identity of the metal). Known Mass of ethanol = 1.55 g Molar mass of ethanol = 46.1 g/mol Mass of water = 200 g c p water = 4.18 J/g o C Temperature increase = 55 o C Unknown Step 2: Solve. (Note: You should find that the specific heat is close to that of two different metals. The work, w, is positive if it is done on the system and negative if it is done by the system. You will find a table of standard enthalpies of formation of many common substances in Appendix G. These values indicate that formation reactions range from highly exothermic (such as 2984 kJ/mol for the formation of P4O10) to strongly endothermic (such as +226.7 kJ/mol for the formation of acetylene, C2H2). It says that 2 moles of of $\ce{CH3OH}$ release $\text{1354 kJ}$. Calculate \({\bf{\Delta H}}_{{\bf{298}}}^{\bf{0}}\)for this reaction and for the condensation of gaseous methanol to liquid methanol. (This amount of energy is enough to melt 99.2 kg, or about 218 lbs, of ice.). Enthalpies of combustion for many substances have been measured; a few of these are listed in Table 5.2. Calculate the molar heat of combustion. That is, the equation in the video and the one above have the exact same value, just one is per mole, the other is per 2 mols of acetylene. The species of algae used are nontoxic, biodegradable, and among the worlds fastest growing organisms. And we're gonna multiply this by one mole of carbon-carbon single bonds. Using enthalpies of formation from T1: Standard Thermodynamic Quantities calculate the heat released when 1.00 L of ethanol combustion. This material has bothoriginal contributions, and contentbuilt upon prior contributions of the LibreTexts Community and other resources,including but not limited to: This page titled 5.7: Enthalpy Calculations is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Robert Belford. Molar enthalpies of formation are intensive properties and are the enthalpy per mole, that is the enthalpy change associated with the formation of one mole of a substance from its elements in their standard states. To begin setting up your experiment you will first place the rod on your work table. Also, these are not reaction enthalpies in the context of a chemical equation (section 5.5.2), but the energy per mol of substance combusted. Also notice that the sum This is the same as saying that 1 mole of of $\ce{CH3OH}$ releases $\text{677 kJ}$. Reactants \(\frac{1}{2}\ce{O2}\) and \(\frac{1}{2}\ce{O2}\) cancel out product O2; product \(\frac{1}{2}\ce{Cl2O}\) cancels reactant \(\frac{1}{2}\ce{Cl2O}\); and reactant \(\dfrac{3}{2}\ce{OF2}\) is cancelled by products \(\frac{1}{2}\ce{OF2}\) and OF2. change in enthalpy for our chemical reaction, it's positive 4,719 minus 5,974, which gives us negative 1,255 kilojoules. Ethanol (CH 3 CH 2 OH) has H o combustion = -326.7 kcal/mole. We can look at this as a two step process. (Note that this is similar to determining the intensive property specific heat from the extensive property heat capacity, as seen previously.). The specific heat Cp of water is 4.18 J/g C. Delta t is the difference between the initial starting temperature and 40 degrees centigrade. The answer is the experimental heat of combustion in kJ/g. a carbon-carbon bond. Accessibility StatementFor more information contact us atinfo@libretexts.orgor check out our status page at https://status.libretexts.org. As discussed, the relationship between internal energy, heat, and work can be represented as U = q + w. Internal energy is an example of a state function (or state variable), whereas heat and work are not state functions. This is a consequence of enthalpy being a state function, and the path of the above three steps has the same energy change as the path for the direct hydrogenation of ethylene. By their definitions, the arithmetic signs of V and w will always be opposite: Substituting this equation and the definition of internal energy into the enthalpy-change equation yields: where qp is the heat of reaction under conditions of constant pressure. 1molrxn 1molC 2 H 2)(1molC 2 H 26gC 2 H 2)(4gC 2 H 2) H 4g =200kJ U=q+w U 4g =200,000J+571.7J=199.4kJ!!! Note: If you do this calculation one step at a time, you would find: As reserves of fossil fuels diminish and become more costly to extract, the search is ongoing for replacement fuel sources for the future. OpenStax is part of Rice University, which is a 501(c)(3) nonprofit. This is also the procedure in using the general equation, as shown. mole of N2 and 1 mole of O2 is correct in this case because the standard enthalpy of formation always refers to 1 mole of product, NO2(g). And since we're Step 3: Combine given eqs. Use the formula q = Cp * m * (delta) t to calculate the heat liberated which heats the water. Use the following enthalpies of formation to calculate the standard enthalpy of combustion of acetylene, #"C"_2"H"_2#. So next, we're gonna where #"p"# stands for "products" and #"r"# stands for "reactants". However, we're gonna go This is usually rearranged slightly to be written as follows, with representing the sum of and n standing for the stoichiometric coefficients: The following example shows in detail why this equation is valid, and how to use it to calculate the enthalpy change for a reaction of interest. How much heat is produced by the combustion of 125 g of acetylene? Direct link to JPOgle 's post An exothermic reaction is.