The enthalpy change for a balanced chemical reaction as written may be calculated by subtracting the sum of the enthalpies of formation for the products from the sum of the enthalpies of formation for the reactants. Because the enthalpies of formation are given per mole, they each need to be multiplied by the stoichiometric coefficients from the balanced chemical equation!ΔH°rxn = Σ [ΔHf°(products)] – Σ [ΔHf°(reactants)]Table of Enthalpies of formation (kJ/mol) Formula Name ΔH°f (kJ/mol) CaCO3(s) calcium carbonate -1207.6 CaO(s) calcium oxide -635.1 CO2(g) carbon dioxide -393.51 C3H5(NO3)3(l) nitroglycerin -364.0 N2(g) nitrogen 0 O2(g) oxygen 0 H2O(l) water -285.83 H2O(g) water -241.82 C3H8(g) propane -103.8 C6H6(l) benzene 49 C6H12O6(s) D-galactose -1286.3 Mg(OH)2(s) magnesium hydroxide -924.66 HF(aq) hydrogen fluoride -332.6 Mg2+(aq) magnseium ion -466.9 F-(aq) fluoride ion -332.6 Using the enthalpies of formation tabulated above, calculate the enthalpy change for the given reactions. CaCO3(s) → CaO(s) + CO2(g) 178.99  kJ/reaction C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(l)   kJ/reaction 4 C3H5(NO3)3(l) → 6 N2(g) + O2(g) + 12 CO2(g) + 10 H2O(g)   kJ/reaction Mg(OH2)(s) + 2 HF(aq) → Mg2+(aq) + 2 F-(aq) + 2 H2O(l)   kJ/reaction C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l)   kJ/reaction Based on your calculations above, how much heat energy would the combustion of 1 mole of solid galactose produce?   kJ Based on your calculations above, how much heat energy would the combustion of 1 mole of gaseous propane produce?   kJ Based on your calculations above, how much heat energy would the decomposition of 1 mole of liquid nitroglycerin produce?   kJ

The enthalpy change for a balanced chemical reaction as written may be calculated by subtracting the sum of the enthalpies of formation for the products from the sum of the enthalpies of formation for the reactants. Because the enthalpies of formation are given per mole, they each need to be multiplied by the stoichiometric coefficients from the balanced chemical equation!ΔH°rxn = Σ [ΔHf°(products)] – Σ [ΔHf°(reactants)]Table of Enthalpies of formation (kJ/mol)

Formula Name ΔH°f (kJ/mol)
CaCO3(s) calcium carbonate -1207.6
CaO(s) calcium oxide -635.1
CO2(g) carbon dioxide -393.51
C3H5(NO3)3(l) nitroglycerin -364.0
N2(g) nitrogen 0
O2(g) oxygen 0
H2O(l) water -285.83
H2O(g) water -241.82
C3H8(g) propane -103.8
C6H6(l) benzene 49
C6H12O6(s) D-galactose -1286.3
Mg(OH)2(s) magnesium hydroxide -924.66
HF(aq) hydrogen fluoride -332.6
Mg2+(aq) magnseium ion -466.9
F(aq) fluoride ion -332.6

Using the enthalpies of formation tabulated above, calculate the enthalpy change for the given reactions.

CaCO3(s) → CaO(s) + CO2(g) 178.99  kJ/reaction

C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(l)   kJ/reaction

4 C3H5(NO3)3(l) → 6 N2(g) + O2(g) + 12 CO2(g) + 10 H2O(g)   kJ/reaction

Mg(OH2)(s) + 2 HF(aq) → Mg2+(aq) + 2 F(aq) + 2 H2O(l)   kJ/reaction

C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l)   kJ/reaction

Based on your calculations above, how much heat energy would the combustion of 1 mole of solid galactose produce?   kJ

Based on your calculations above, how much heat energy would the combustion of 1 mole of gaseous propane produce?   kJ

Based on your calculations above, how much heat energy would the decomposition of 1 mole of liquid nitroglycerin produce?   kJ

A city of 100,000 people uses approximately 1.0 x 1011 kJ of energy per day. Suppose all of that energy comes from the combustion of liquid pentane (C5H12, density = 0.626 g/mL, molar mass = 72.151 g/mol) to form gaseous carbon dioxide and gaseous water. Use the standard enthalpies of formation (in the tables above and below) to determine how much energy is provided from the combustion reaction, and then determine how many gallons of pentane each person would require for their daily energy consumption. Much of this problem can be solved quite readily using dimensional analysis.

Table of Enthalpies of formation (kJ/mol)

Formula Name ΔH°f (kJ/mol)
C5H12(l) pentane -173.5
C6H14(l) hexane -198.8
C7H16(l) heptane -224.2
C8H18(l) octane -250.1
C2H6O (l) ethanol -277.6

 gallons of pentane per person per day.

Tired of numerous paper assignments?
Rely on us and receive professional paper writing assistance!
Professional paper Writing Assistance

Who We Are

We are a professional website for customized writing. If you searched a question and stumbled into our website, you are in the right place to receive assistance with your coursework.

Do you handle any type of coursework?

Yes. We have displayed prior orders to demonstrate our experience. We can answer this question for you as we have previously. Please fill out our Order Form so that we may ensure its flawlessness. Correctly completing the order form will help our staff with reference, requirements, and future communication.

Is it hard to Place an Order?

  1. Click on the “Order Now” tab at the top menu or “GET A FREE QUOTE” icon at the bottom and a new page will appear with an order form to be filled
  2. Fill in the initial requirements in the small order form located on the home page and press “continue” button to proceed to the main order form or press “order” button in the header menu. Starting from there let our system intuitively guide you through all steps of ordering process. Submit detailed paper instructions, upload necessary files if needed and provide your contact information – you are almost done!
  3. Proceed with the payment- click on “PROCEED TO CHECKOUT” at the bottom of the page. From there, the payment sections will show, follow the guided payment process and your order will be available for our writing team to work on it. All your payments are processed securely through PayPal. This enables us to guarantee a 100% security of your funds and process payments swiftly.
  4. Delivery- Once finished, your final paper will be available for download through your personal dashboard. You will also receive an email notification with a copy of your paper attached to it.