Question

Question 1

Which of the following reactions could be used to calculate the standard state heat of formation of the products, if performed in a calorimeter with one mole of each reactant, and performed isothermally at 298 K?

a) 2NO2 (g) → N2O4 (g)

b) 4NH3 (g) + 5O2 (g) → 4NO (g) + 6H2O (g)

c) N2 (g) + 2O2 (g) → 2NO2 (g)

d) CO2 (aq) + H2O (l) → H2CO3 (aq)

e) H2 (g) + O2 (g) → H2O2 (l)

f) Fe (s) → Fe3+(aq)

g) 2O3 (g) + Cl2 (g) → Cl2O6 (g)

h) C (s, graphite) + O2 (g) → CO2 (g)

Regarding the previous question for how standard heat of formation can be initially calculated, would you absolutely need to produce exactly one Mole of product in order to determine the standard state value? Choose the best answer

No, because enthalpy is not a state function, you use any arbitrary value you wish and label it as the standard state per Mole

Yes, because enthalpy is not a state function, any value other than one Mole cannot actually be determined

No, because enthalpy is a state function, as long as you know what fraction of a mole you have produced, you can normalize the value to be per Mole

Yes, because enthalpy is a state function, if it is not a Mole produced, you can never determine the value per Mole

Answer 1

Question 1)

answers :

e) H2 (g) + O2 (g) → H2O2 (l)

h) C (s, graphite) + O2 (g) → CO2 (g)

explanation:

standard state heat of formation of the products should be for one mole and it should be formed from elements from their stnadard states only.

among given all one these are correct because one mole product formed and these are from standard state reactants

next

question 2)

Yes, because enthalpy is a state function, if it is not a Mole produced, you can never determine the value per Mole