For which one of the following reactions is δh°rxn equal to the heat of formation of the product? for which one of the following reactions is h°rxn equal to the heat of formation of the product? (1/2)a. n2 (g) + o2 (g) → no2(g) b. p (g) + 4h (g) + br (g) → ph4br (l)c. 12c (g) + 11h2 (g) + 11o (g) → c6h22o11 (g) d. n2 (g) + 3h2 (g) → 2nh3 (g) e. 6c (s) + 6h (g) → c6h6 (l)

The reaction in which ΔH°rxn is equal to the heat of formation of the product is 1/2N2 (g) + O2 (g) → NO2(g)

The standard enthalpy change for a reaction is obtained as the difference between the enthalpy change of the reactants and the products. It is a state function hence it depends on the initial and final states of the system.

Note that the standard enthalpy change for a reaction is calculated from the formula;

ΔH°rxn = [n∑(ΔH°products - n∑(ΔH°reactants]

For the reaction; 1/2N2 (g) + O2 (g) → NO2(g), the ΔH°reactants = zero because they are in standard state therefore, ΔH°rxn is equal to the heat of formation of the product.

Learn more:  link

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g

14,16 18 reacts with group