37.2 liters of a gas has a pressure of 362.43 kPa at 46.5 °C. If the pressure increases to 693.9 kPa and the temperature to 149.2°C, what would be the new volume of the gas?
Select one:
O a. 25.7
O b. 62.3
O c. 54
O d. 25.672

d. temperate deciduous forest

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 :