Using this reversible reaction, answer the questions below: n2o4 2no2 (colorless) (reddish-brown) -as the temperature increased, what happened to the n2o4 concentration? -was the formation of reactants or products favored by the addition of heat? -which reaction is exothermic? right to left or left to right? -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

1) As the temperature increased what happened to the N2O4 concentration, it decreased

2) Formation of products, products are the right hand side of the equation.

3) Right to left is exothermic

4) Change in enthalpy N2O4 ⇔2NO2, HR = -14 kcal

As it's an exothermic reaction, the enthalpy of the reactants is higher than the products and the sign of the HR will be negative.

Explanation:

N2O4 ⇔2NO2

(colorless) (reddish-brown)

1) As the temperature increased what happened to the N2O4 concentration, it decreased

2) Formation of products, products are the right hand side of the equation.

3) Right to left is exothermic

4) Change in enthalpy N2O4 ⇔2NO2, HR = -14 kcal

As it's an exothermic reaction, the enthalpy of the reactants is higher than the products and the sign of the HR will be negative.

the answer would be b i believe