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)
Question
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)
Solution
The standard heat of reaction, also known as the enthalpy change of reaction, is calculated using the formula:
ΔH°reaction = Σ ΔH°f(products) - Σ ΔH°f(reactants)
where ΔH°f is the standard enthalpy of formation of the substances.
For the given reaction:
3 CH4(g) + 4 O3(g) -> 3 CO2(g) + 6 H2O(g)
The standard enthalpy change of the reaction would be calculated as follows:
ΔH°reaction = [3 x ΔH°f(CO2(g)) + 6 x ΔH°f(H2O(g))] - [3 x ΔH°f(CH4(g)) + 4 x ΔH°f(O3(g))]
Substituting the given values into the equation:
ΔH°reaction = [3 x -394 kJ/mol + 6 x -242 kJ/mol] - [3 x -74.8 kJ/mol + 4 x 143 kJ/mol]
So, the correct answer is (c).
Similar Questions
The bond enthalpies for six selected chemical bonds are shown below.Bond Bond enthalpyΔH°298 (kJ/mol)H―H 436C―H 413C=O 532N―H 335O―H 499O=O 495For a reaction in which 2 moles of H2(g) reacts with 1 mole of O2(g) to form 2 moles of H2O(g), what is the heat of the reaction?A.1367 kJB.369 kJC.−629 kJD.−1996 kJ
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
During a reaction, the enthalpy of formation of an intermediate is 90.3 kJ/mol. During the reaction, 2 moles of the intermediate are formed as a reactant. What is the enthalpy value for this step of the reaction?A.-90.3 kJB.90.3 kJC.180.6 kJD.-180.6 kJSUBMITarrow_backPREVIOUS
Which would not have a heat of formation of 0 kJ/mol? Group of answer choicesF2(g)H2(g)S2(g)Kr(g)
Calculate the heat of the following reaction:2C2H2 + 5O2 → 4CO2 + 2H2OGiven the following heats of formation:C2H2 ΔH°f= 226.7 kJ/molO2 ΔH°f= 0.0 kJ/molCO2 ΔH°f= -393.5 kJ/molH2O ΔH°f= -241.8 kJ/molGroup of answer choices0.0 kJ/mol-2511 kJ/mol-2284.3 kJ/mol408.6 kJ/mol
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.