Ozone (o3) in the atmosphere can be converted to oxygen gas by reaction with nitric oxide (no). nitrogen dioxide is also produced in the reaction. what is the enthalpy change when 8.50l of ozone at a pressure of 1.00 atm and 25°c reacts with 12.00 l of nitric oxide at the same initial pressure and temperature? [δh°f (no) = 90.4 kj/mol; δh°f (no2) = 33.85 kj/mol; δh°f (o3) = 142.2 kj/mol]

Ozone (O3) in the atmosphere can be converted to oxygen gas by reaction with nitric oxide (NO). Nitrogen dioxide is also produced in the reaction. What is the enthalpy change when 8.50L of ozone at a pressure of 1.00 atm and 25°C reacts with 12.00 L of nitric oxide at the same initial pressure and temperature? [ΔH°f (NO) = 90.4 kJ/mol; ΔH°f (NO2) = 33.85 kJ/mol; ΔH°f (O3) = 142.2 kJ/mol]

(A) - 69.2 kJ

(B) - 19.7 kJ

(C) - 1690 kJ

(D) - 97.6 kJ

(E) - 167 kJ

The enthalpy of change of the reaction = - 68.966 kJ ≅69.0kJ

The best option is (A) - 69.2 kJ ≅ - 69.0kJ

Explanation:

From ideal gas law, P x V = n x R x T

calculate the moles of NO and O3. and find the limiting reactant

NO + O3 - -> O2 + NO2

Finding the difference of the sum between the sum of the enthalpies of the products and that of the reactants according to the enthalpyy of formation equation and taking note that he enthalpy of formation of ΔH°f (O2) = 0 kJ/mol

From the chemical reaction equation, 1 mole of O3 reacts with 1 mole of NO to produce 1 mole of O2 and 1 mole of NO2

from the ideal gas equation, P x V=n x R x T

Where P = Gas pressure

V = Volume

n = Number of moles

R = Universal gas constant

We then have for ozone, O3

1 * 8.5 = n * 0.08205*298.15

Therefore, n = 8.5/24.5 = 0.347 moles of O3

Similarly, the number of moles of NO is given by

1 * 12 = n * 0.08205*298.15

Therefore the number of moles of NO in the reaction = 0.49 moles of NO

Since 1 mole of O3 can only react with 1 mole NO then the O3 is the limiting reagent

Hence the number of moles of reactants and products is 0.347 moles each

The enthalpy of formation equation states that the standard enthalpy change of formation is the difference between the sum total of the enthalpies of formation of the products and the sum total of the enthalpies of formation of the reactants

Therefore, we multiply each of the above enthalpies of formation by the number of moles which take place in the reaction, in this case, 0.347 moles

thus

For NO = 90.4 kj/mol the heat of formation of 0.347 moles 90.4*0.347 = 31.369 kJ

Similarly, NO2 heat of formation of 0.347 moles = 33.85*0.347 = 11.746kJ and that of O3 = 49.343kJ

Note the enthalpy of formation of O2 = 0 kJ

Thus we have the enthalpy of change of the reaction given as

11.746kJ - (49.343kJ + 31.369 kJ) = - 68.966 kJ

The enthalpy of change of the reaction = - 68.966 kJ ≅69 kJ