Using this reversible reaction, answer the following:
n2o4 < --> 2no2
(colorless) (reddish-brown)

-as the temperature increased, what happened to the [n2o4]?

-was the formation of reactants or products favored by the addition of heat?

-which reaction is exothermic?

-if the change of enthalpy of this reaction when proceeding left to right is 14 kcal, which chemical equation is correct?

n2o4 < --> 2no2 + 14 kcal
n2o4 < --> 2no2, hr = +14 kcal
n2o4 + 14 kcal < --> 2no2
n2o4 < --> 2no2, hr = -14 kcal

Question 1.

The concentration of [N2O4] will decrease.

Explanation:As the reverse reaction which is the formation of N2O4 is an exothermic reaction and the formation of NO2 is an endothermic reaction.Hence heating will favor the forward reaction and stops the reverse reaction.N2O4 --> 2NO2So the concentration of N2O4 decreases.Question: 2.

The formation of the products is favored by the addition of heat.

Explanation:As the reverse reaction which is the formation of N2O4 is an exothermic reaction and the formation of NO2 is an endothermic reaction.So the addition of heat will favor endothermic reaction.Hence the formation of NO2 will be favored which is a product.Question no.4:

The reverse reaction is an exothermic reaction.

N2O4 <-- 2NO2.

Explanation:N2O4 which is a reactant in this reaction is colorless compound.Its breakdown absorbs energy and its formation release energy in form of heat.So the reverse reaction which releases energy is an exothermic reaction.Question no. 4

The correct option is D which is N2O4 <--> 2NO2, HR = -14 kcal.

Explanation:Enthalpy is the change in energy in reactants and products.It is actually the difference in energies of products and reactant.As the enthalpy change from left to right, it means reactants have internal energy 14 kcal high than products and it will be released.And in option D which is shown as HR = -14 kcal.

Google it that might work