(b) Calculate the standard free energy change of the reaction: p =-145,000 Cal/mole and JG1000, M20 > = -55,000 Cal/mole. Also comment on the possibility of reducing Moo, by H2 at 727 at 1 atm pressure.

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 :

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