Acetylene is used in blow torches, and burns according to the following
equation:
2 C2H2(g) + 5 O2(g) → 4 CO2(g) + 2 H2O(g)
Use the following information to calculate the heat of reaction:?

Question

Acetylene is used in blow torches, and burns according to the following
equation:
2 C2H2(g) + 5 O2(g) → 4 CO2(g) + 2 H2O(g)
Use the following information to calculate the heat of reaction:

Hfo
(H2O(g))= -241.82 kJ/mol

Hfo
(CO2(g))= -393.5 kJ/mol

Hfo
(C2H2(g))=226.77 kJ/mol

Luna Chen · Accepted Answer

#Delta"H"_(rxn)=-2511.04" ""kJ"#

Hess' Law states that the overall enthalpy change of a reaction is independent of the route taken.

Thermodynamics is concerned with initial and final states and the law is a consequence of the conservation of energy.

You can solve this problem by constructing a Hess Cycle.

Write down the reaction you are interested in. Below this write down the elements from which the reactants and products are made.

Then complete the cycle as shown:

Notice I have multiplied the #Delta"H"_"f"# values by the relevant stoichiometric numbers.

In energy terms the #color(blue)"BLUE"# route must equal the #color(red)"RED"# route since the arrows start and finish in the same place.

So we can write:

#(2xx226.77)+Delta"H"=(4xx-393.5)+(2xx-241.82)#

#:.453.4+Delta"H"=-1574-483.64#

#Delta"H"=-2511.04" ""kJ"#

Jeremiah Adams · Answer

undefined To calculate the heat of reaction ($ΔH$), we need to use the standard enthalpies of formation for the substances involved in the reaction. The standard enthalpy of formation for a substance is the change in enthalpy when one mole of the substance is formed from its elements in their standard states.

Given:
- Standard enthalpy of formation of CO₂ = -393.5 kJ/mol
- Standard enthalpy of formation of H₂O = -241.8 kJ/mol
- Standard enthalpy of formation of C₂H₂ = 226.7 kJ/mol

Using the given equation:

2 C₂H₂(g) + 5 O₂(g) → 4 CO₂(g) + 2 H₂O(g)

The heat of reaction ($ΔH$) can be calculated as:

$$ΔH = \sumΔH_{\text{products}} - \sumΔH_{\text{reactants}}$$

$$ΔH = [4(-393.5) + 2(-241.8)] - [2(226.7) + 5(0)]$$

$$ΔH = [-1574 - 483.6] - [453.4]$$

$$ΔH = -2057.6 - 453.4$$

$$ΔH = -2511 \, \text{kJ/mol}$$

Therefore, the heat of reaction for the combustion of acetylene is -2511 kJ/mol.

Acetylene is used in blow torches, and burns according to the following equation: 2 C2H2(g) + 5 O2(g) → 4 CO2(g) + 2 H2O(g) Use the following information to calculate the heat of reaction:?

Acetylene is used in blow torches, and burns according to the following equation: 2 C2H2(g) + 5 O2(g) → 4 CO2(g) + 2 H2O(g) Use the following information to calculate the heat of reaction: Hfo (H2O(g))= -241.82 kJ/mol Hfo (CO2(g))= -393.5 kJ/mol Hfo (C2H2(g))=226.77 kJ/mol

Acetylene is used in blow torches, and burns according to the following
equation:
2 C2H2(g) + 5 O2(g) → 4 CO2(g) + 2 H2O(g)
Use the following information to calculate the heat of reaction:

Hfo
(H2O(g))= -241.82 kJ/mol

Hfo
(CO2(g))= -393.5 kJ/mol

Hfo
(C2H2(g))=226.77 kJ/mol