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Eddi Din [679]
2 years ago
9

Use the reaction and bond information to answer the question. H2 CO2 → CH2O2 Reactant bond energies: H–H is 432 kJ/mol, C=O is 7

99 kJ/mol Product bond energies: C–H is 413 kJ/mol, C=O is 745 kJ/mol, C–O is 358 kJ/mol, O–H is 467 kJ/mol How much energy must this system absorb before the reaction can begin?.
Chemistry
1 answer:
Anna11 [10]2 years ago
7 0

The energy absorbed by the system before the reaction start has been 752 kJ/mol.

The bond energy has been defined as the energy possessed by the bond that has been converted into the heat energy and utilized in the reaction.

The energy has been conserved in a chemical reaction. Thus, the energy absorbed by the reactant in the reaction has been the difference in the energy of the reactant and product.

<h3>Computation for the energy absorbed</h3><h3 />

The given reaction has been:

\rm H_2\;+\;CO_2\;\rightarrow\;CH_2O_2

The bond energy of the reactant has been:

H-H bond energy = 432 kJ/mol

2 C=O bond energy= 799 kJ/mol

The total energy of the reactant has been 432 kJ/mol + 799 kJ/mol

The total energy of the reactant has been 1231 kJ/mol

The bond energy of the product has been:

C-O bonds = 358 kJ/mol

C=O bonds = 745 kJ/mol

1 C-H bond = 413 kJ/mol

1 O-H bond = 467 kJ/mol

The total energy of the product has been 358 + 745 + 413 + 467 kJ/mol

The total energy of the product has been 1,983 kJ/mol

According to the law of conservation of energy:

\text{Product energy=Reactant energy + energy absorbed}

Substituting the values for energy absorbed:

\text{1,983 kJ/mol=1231\;kJ/mol+ energy absorbed}\\\rm Energy\;absorbed=1983-1231\;kJ/mol\\Energy\;absorbed=752\;kJ/mol

The energy absorbed by the system before the reaction start has been 752 kJ/mol.

Learn more about bond energy, here:

brainly.com/question/866298

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The metabolic oxidation of glucose, C6H12O6, in our bodies produces CO2, which is expelled from our lungs as a gas.
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Answer:

\large \boxed{\text{21.6 L}}

Explanation:

We must do the conversions

mass of C₆H₁₂O₆ ⟶ moles of C₆H₁₂O₆ ⟶ moles of CO₂ ⟶ volume of CO₂

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ:        180.16

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m/g:      24.5

(a) Moles of C₆H₁₂O₆

\text{Moles of C$_{6}$H$_{12}$O}_{6} = \text{24.5 g C$_{6}$H$_{12}$O}_{6}\times \dfrac{\text{1 mol C$_{6}$H$_{12}$O}_{6}}{\text{180.16 g C$_{6}$H$_{12}$O}_{6}}\\\\= \text{0.1360 mol C$_{6}$H$_{12}$O}_{6}

(b) Moles of CO₂

\text{Moles of CO}_{2} =\text{0.1360 mol C$_{6}$H$_{12}$O}_{6} \times \dfrac{\text{6 mol CO}_{2}}{\text{1 mol C$_{6}$H$_{12}$O}_{6}} = \text{0.8159 mol CO}_{2}

(c) Volume of CO₂

We can use the Ideal Gas Law.

pV = nRT

Data:

p = 0.960 atm

n = 0.8159 mol

T = 37  °C

(i) Convert the temperature to kelvins

T = (37 + 273.15) K= 310.15 K

(ii) Calculate the volume

\begin{array}{rcl}pV &=& nRT\\\text{0.960 atm} \times V & = & \text{0.8159 mol} \times \text{0.082 06 L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{310.15 K}\\0.960V & = & \text{20.77 L}\\V & = & \textbf{21.6 L} \\\end{array}\\\text{The volume of carbon dioxide is $\large \boxed{\textbf{21.6 L}}$}

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