what is the enthalpy of formation for c4h8??????????????

Expert Solution

samet mamat answered 2 years ago

standard enthalpy change of formation

a. Define standard enthalpy change of formation.b. Calculate the standard enthalpy change of formation of methane from the following standard enthalpy changes of combustion:carbon = –394 kJ mol–1hydrogen = –286 kJ mol–1methane = –891 kJ mol–1c. Calculate the standard enthalpy change of combustion of methane using the following bond energies:E(C — H) = +412 kJ mol–1E(O — O) = +496 kJ mol–1E(C — O) = +805 kJ mol–1E(O — H) = +463 kJ mol–1

Calculate the enthalpy of oxidation per mole for N2 and for C2H2 (the enthalpy of formation...

Calculate the enthalpy of oxidation per mole for N2 and for C2H2 (the enthalpy of formation of N2O5(g) is 11.30 kJ/mol).

For the formation of Silica: Si(s) + O2 --> SiO2(s) Find the enthalpy of formation for...

For the formation of Silica: Si(s) + O2 --> SiO2(s) Find the enthalpy of formation for SiO2 at 500 K,1 atm (in symbols and in numbers) _________________________________________________________________________ Find the equilibrium constant for ther above reaction at 500K (in symbols & in numbers).

In which of the reactions below is ΔH not an enthalpy of formation? Group of answer...

In which of the reactions below is ΔH not an enthalpy of formation? Group of answer choices Both b and c are not enthalpy of formation reactions. 2C(s) + O2(g) → 2CO(g) Mg(s) + Cl2(g) → MgCl2(s) Ca(s) + 1/2O2(g) → CaO(s) C(s) + O2(g) → CO2(g)

show me step by step calculations on enthalpy of formation of hydrogen peroxide

A.) the standard enthalpy of formation of the reaction C2H4(g)+H2O(l)=C2H5OH(l) B.) Use standard enthalpies of formation...

A.) the standard enthalpy of formation of the reaction C2H4(g)+H2O(l)=C2H5OH(l) B.) Use standard enthalpies of formation to calculate the standard enthalphy change of: the reaction of methane gas, CH4, with chlorine rine to form liquid chloroform, CHCl3. Gaseuous hydrogen chloridee is the other product. C.) Use standard enthalpies of formation to calculate the standard enthalphy change of PCl3(g)+HCl(g)=PCl5(g)+H2(g)

For a butanol combustion process (formation enthalpy -332,000 kJ / kmol) C4H10O with an excess of...

For a butanol combustion process (formation enthalpy -332,000 kJ / kmol) C4H10O with an excess of 25% air (reagents are at standard conditions T = 25 C, P = 1atm). Determine: a. Adiabatic flame temperature b. If the products come out at a temperature of 600 K, determine the heat generated in the combustion chamber.

Use bond-energy data to calculate the enthalpy of formation for eah of the following compounds at...

Use bond-energy data to calculate the enthalpy of formation for eah of the following compounds at 25 C a) n-Octane, C8H18(g) b) Napthalene, C10H8(g) c) Formaldehyde, H2CO(g) d) Formic acid, HCOOH(g) Give the most likely reason for the largest discrepancies between your calculated values and the ones in the tables.

1. Calculate the enthalpy change, ΔH, for the formation of 1 mole of strontium carbonate (the...

1. Calculate the enthalpy change, ΔH, for the formation of 1 mole of strontium carbonate (the material that gives the red color in fireworks) from its elements. Sr(s) + C(s) + 1.5 O2(g) → SrCO3(s) Use the following information: 2 Sr(s) + O2(g) → 2 SrO(s) ΔH = -1184 kJ SrO(s) + CO2(g) → SrCO3(s) ΔH = -234 kJ C(s) + O2(g) → CO2(g) ΔH = -394 kJ 2. Write chemical reactions that describe the formation of the following compounds:...

Consider aluminum and oxygen. In their natural states, their standard enthalpy of formation (i.e., the energy...

Consider aluminum and oxygen. In their natural states, their standard enthalpy of formation (i.e., the energy of formation at RTP) is zero. Every kilogram of aluminum has (at RTP) an entropy of 1.05 kJ/K, whereas every kilogram of oxygen has an entropy of 6.41 kJ/K. Aluminum burns fiercely, forming an oxide (Al2O3) and releasing energy. The standard enthalpy of formation of the oxide is –1.67 GJ/kmol. The entropy of the oxide is 51.0 kJ/K per kilomole. According to the second...

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