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]

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

a gas, having a molecular weight of 13 kg/kmol and a kinematic viscosity of 0.25 cm2/s, flows through a pipe 0.25 m internal diameter and 5 km long at the rate of 0.4 m3/s and is delivered at atmospheric pressure.