To find the enthalpy of a reaction using heats of formation…
Question
To find the enthalpy of a reaction using heats of formation…
Solution
Sure, here are the steps to find the enthalpy of a reaction using heats of formation:
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Write down the balanced chemical equation for the reaction.
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Look up the standard heats of formation for each reactant and product in the reaction. These values are usually given in tables in chemistry textbooks or can be found online. The standard heat of formation is the amount of energy required to form one mole of a substance from its elements in their standard states.
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Multiply the standard heat of formation for each reactant and product by the number of moles of that substance in the balanced chemical equation.
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Sum up the total standard heats of formation for all the reactants and for all the products.
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Subtract the sum of the standard heats of formation of the reactants from the sum of the standard heats of formation of the products. This gives you the enthalpy change for the reaction.
The formula is: ΔH = Σ ΔHf (products) - Σ ΔHf (reactants)
Where:
- ΔH is the enthalpy change of the reaction
- ΔHf is the standard heat of formation
- The Σ symbol means "the sum of"
Similar Questions
To find the enthalpy of a reaction using heats of formation…Group of answer choicessubtract the product heats of formation from reactant heats of formationsubtract the reactant heats of formation from product heats of formation
Given the following enthalpies of formation in kJ/mol:CH4(g) = - 74.8 O3(g) = 143H2O(g) = - 242 CO2(g) = - 394what is the standard heat of reaction in kJ/mol for the reaction below?3 CH4(g) + 4 O3(g) 3 CO2(g) + 6 H2O(g)Question 6Select one:a.(3 x -394 + 6 x -242) + ( 3 x -74.8 + 4 x 143) b.(3 x -394 + 6 x -242) - ( 3 x 74.8 + 4 x -143) c.(3 x -394 + 6 x -242) - ( 3 x -74.8 + 4 x 143) d.( 3 x -74.8 + 4 x 143) - (3 x -394 + 6 x -242)e.( 3 x -74.8 + 4 x 143) + (3 x -394 + 6 x -242)
Calculate the heat of the following reaction:C6H12O6 → 2C2H5OH + 2CO2Given the following heats of formation:C6H12O6 ΔH°f= -1260 kJ/molC2H5OH ΔH°f= -277.7 kJ/molCO2 ΔH°f= -393.5 kJ/molGroup of answer choices-336.1 kJ/mol-1266 kJ/mol-22.6 kJ/mol-82.4 kJ/mol
Given the following thermochemical data, what is the enthalpy of formation ΔfHo for C2H5OH(l) C(s) + O2(g) → CO2(g) ΔHo = –393 kJ mol–1H2(g) + 1/2 O2(g) → H2O (g) ΔHo = -298 kJ mol–1C2H5OH +3O2 → 3H2O (g) + 2CO2(g) ΔHo =–1369 kJ mol–1 Use-full formula ΔrxnHo= ∑ ΔfHoproducts - ∑ ΔfHoreactants
Consider the gas-phase reaction below.2 H2S(g) + CH4(g) → 4 H2(g) + CS2(g)Which expression can be used to calculate the heat of reaction (ΔH°rxn) for the reaction shown using bond enthalpies (ΔH°bond)?A.ΔH°rxn=(2×ΔH°S—H)+ΔH°C—H−(4×ΔH°H—H)−ΔH°C=S∆𝐻°rxn=2×∆𝐻°S―H+∆𝐻°C―H-4×∆𝐻°H―H-∆𝐻°C=SB.ΔH°rxn=(4×ΔH°S—H)+(4×ΔH°C—H)−(4×ΔH°H—H)−(2×ΔH°C=S)∆𝐻°rxn=4×∆𝐻°S―H+4×∆𝐻°C―H-4×∆𝐻°H―H-2×∆𝐻°C=SC.ΔH°rxn=(4×ΔH°H—H)+(2×ΔH°C=S)−(4×ΔH°S—H)−(4×ΔH°C—H)∆𝐻°rxn=4×∆𝐻°H―H+2×∆𝐻°C=S-4×∆𝐻°S―H-4×∆𝐻°C―HD.ΔH°rxn=(4×ΔH°H—H)+(2×ΔH°C=S)+(4×ΔH°S—H)+(4×ΔH°C—H)∆𝐻°rxn=4×∆𝐻°H―H+2×∆𝐻°C=S+4×∆𝐻°S―H+4×∆𝐻°C―HSubmit
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