Question

What is the enthalpy for reaction 1 reversed? reaction 1 reversed: 2CO2+3H2O→C2H5OH+3O2?

Answer 1

The enthalpy for the reaction ${{\text{C}}_2}{{\text{H}}_5}{\text{OH}}+{\text{3}}{{\text{O}}_2}\to{\text{2C}}{{\text{O}}_2}+3{{\text{H}}_2}{\text{O}}$is $\boxed{1368\;{\text{kJ}}}$.

Further Explanation:

This problem is based upon Hess’s Law. Hess’s law is utilized for calculating the enthalpy change of a reaction that can be obtained simply by summation of two or more reactions. In accordance with the Hess’s law, ${\Delta H}}$of the overall reaction is obtained by adding the enthalpy change for each individual step reaction involved to obtain the overall reaction.

$\boxed{{\Delta }}{{\text{H}}_{{\text{overall rxn}}}}={\Delta }}{{\text{H}}_{1}}}+{\Delta }}{{\text{H}}_{\text{2}}} + ....... +{\Delta }}{{\text{H}}_{\text{n}}}}$

Enthalpy is defined as a state function and therefore its value depends upon the initial and final state of the system but not upon the path. This is the reason that the overall reaction can be simply obtained by adding or subtracting the enthalpy change of the individual steps utilized to get the final reaction.

Combustion reactions:

These are the reactions that take place when hydrocarbons are burnt in the presence of oxygen to form carbon dioxide and water. These are also referred to as burning.

Example of combustion reactions are as follows:

(a) ${\text{C}}{{\text{H}}_4}+{{\text{O}}_2}\to{\text{C}}{{\text{O}}_2}+{{\text{H}}_2}{\text{O}}$

(b) ${{\text{C}}_{10}}{{\text{H}}_{14}}+12{{\text{O}}_2}\to10{\text{C}}{{\text{O}}_2}+4{{\text{H}}_2}{\text{O}}$

The given reaction is as follows:

${\text{2C}}{{\text{O}}_2}+3{{\text{H}}_2}{\text{O}}\to{{\text{C}}_2}{{\text{H}}_5}{\text{OH}}+{\text{3}}{{\text{O}}_2}$

The value of ${\Delta }}{{\text{H}}_{\text{1}}}$is -1368 kJ.

Step 1: The enthalpy change of the following reaction is ${\Delta }}{{\text{H}}_{\text{1}}}$.

${\text{2C}}{{\text{O}}_2}+3{{\text{H}}_2}{\text{O}}\to{{\text{C}}_2}{{\text{H}}_5}{\text{OH}}+{\text{3}}{{\text{O}}_2}$                                      …… (1)

The reverse reaction for the given reaction is as follows:

${{\text{C}}_2}{{\text{H}}_5}{\text{OH}}+{\text{3}}{{\text{O}}_2}\to{\text{2C}}{{\text{O}}_2}+3{{\text{H}}_2}{\text{O}}$

Step 2: The enthalpy change of the following reaction is ${\Delta }}{{\text{H}}_{\text{2}}}$.

${{\text{C}}_2}{{\text{H}}_5}{\text{OH}}+{\text{3}}{{\text{O}}_2}\to{\text{2C}}{{\text{O}}_2}+3{{\text{H}}_2}{\text{O}}$                                       …… (2)

The reaction (2) can be obtained by reversing the reaction (1) so the value of ${\Delta }}{{\text{H}}_{\text{2}}}$ can be obtained as follows:

$\begin{gathered}\Delta{{\text{H}}_2}=-\Delta{{\text{H}}_{\text{1}}}\\=-\left({1368\;{\text{kJ}}}\right)\\=1368\;{\text{kJ}}\\\end{gathered}$

Learn more:

1. Dissociation of ionic compounds: brainly.com/question/5425813

2. Find the enthalpy of dissociation of 1 mole of MgO: brainly.com/question/2416245

Answer details:

Grade: Senior school

Subject: Chemistry

Chapter: Thermodynamics

Keywords: Hess’s Law, enthalpy, enthalpy change, CO2, H2O, O2, C2H5OH, 1368 kJ, -1368 kJ, state function, initial, final, reverse reaction, reversing, 2CO2, 3H2O.

Answer 2

Answer is: the enthalpy for reaction 1 reversed is 1370 kJ/mol.

Balanced chemical reaction 1:
C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O  ΔrH= -1370 kJ/mol.
Reaction 1 is exothermic reaction (heat is released), than reversed reaction is endothermic (chemical reaction that absorbs more energy than it releases) and has same value for enthalpy, but positive sign.