Use the following data to calculate the standard enthalpy of formation of pentane, C5H12(l)?
C5H12(l) + 8 O2 (g) -> 5 CO2 (g) + 6 H2O (l) delta H=-3535.9 kJ/mol

delta Hf of CO2 (g) = -393.5 kJ/mol

delta Hf of H2O (l) = -285.8 kJ/mol

please show all work

Explanation:

C₅H₁₂(l) + 8O₂(g) ⟶ 5CO₂(g) + 6H₂O(l); ΔᵣH = -3535.9 kJ·mol⁻¹

The formula for calculating the enthalpy change of a reaction by using the enthalpies of formation of reactants and products is

Let x = the heat of formation of pentane

                            C₅H₁₂(l) + 8O₂(g) ⟶ 5CO₂(g) + 6H₂O(l)

ΔH°f/kJ·mol⁻¹:            x                          -393.5      -285.8

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

1. a)c-14 has two more protons

2. d)they have different numbers of protons, neutrons, and valence electrons; they are all gases at room temperature.

3. c) iron and nickel.