Suppose we have a 12.2 l sample containing 0.50 mol oxygen gas at a pressure of 1 atm and a temperature of 25 degrees celcius. if all this oxygen were converted to ozone at the same temperature and pressure, what would be the volume of the ozone? balanced equation: 3o2 → 2o3a. 6.5lb. 2.8lc. 8.1ld. 7.4l

c. 8.1 L

Explanation:

Given that:-

Moles of oxygen gas = 0.50 mol

According to the reaction shown below as:-

3 moles of oxygen gas on reaction gives 2 moles of ozone

Also,

1 mole of oxygen gas on reaction gives 2/3 moles of ozone

So,

0.50 mole of oxygen gas on reaction gives moles of ozone

Moles of ozone = 0.3333 mol

Pressure = 1 atm

Temperature = 25.0 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (25.0 + 273.15) K = 298.15 K  

Volume = ?

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

1 atm × V = 0.3333 mol × 0.0821 L.atm/K.mol × 298.15 K  

⇒V = 8.1 L

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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