Draw the major monobromination product when the following alkane is subjected to radical bromination at 25 °c. if there is more than one product

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 :

if the temperature drops and becomes cool

then the atmosphere can be freezing as the water vapor in the air is at maximum (100%) and can be severely cold

also can be snowing lightly as the moisture in the air will start to solidify

and btw this is not chemistry but