00:01
Hi, in this question we have to find out the energy needed to form an o2 product here.
00:09
Now first we have to balance this equation.
00:12
Now balancing nitrogen both sides by adding 2 here and now balancing oxygen both sides by adding 2 here.
00:20
Now we have to firstly calculate the moles of nitrogen and oxygen reacting.
00:26
So for nitrogen using this ideal gas equation we can calculate moles and is equal to pressure into volume divided by r into temperature.
00:40
Now here, r is equal to gas constant, which has value 0 .08 to atmospheric liter mole inverse, kelvin inverse.
00:55
Now, temperature here is 100 degrees celsius for both oxygen and nitrogen.
01:00
So we have to convert it into kelvin, that is 100 plus.
01:06
273 kelvin that is is equal to 373 kelvin now volume is equal to 304 ml clearly one ml is equal to 10 r to per minus 3 from here so this is is equal to 304 into 10 to power minus 3 liters so volume is equal to 0 .304 liter now from here we can calculate n for nitrogen that is moles of nitrogen as p is 3 .32 atmospheric, multiply by volume is 0 .304 liters divided by r value, that is 0 .082, atmospheric liter, mole inverse and kelvin inverse, multiply by the temperature, that is 373 kelvin.
01:58
Now on calculating this, we will get moles of nitrogen as 0 .033 mole.
02:05
Now, finding out the moles of oxygen as, first we have to calculate the volume that is is equal to 410 ml, this will be equal to 410 multiplied by 10 raised to per minus 3 litres, that is is equal to 0 .410 liters.
02:24
Now putting the value in equation that is n is equal to pressure is 3 .32 atmospheric, multiply by the volume that is 0 .410.
02:37
Liters divided by 0 .082 atmospheric liter mole inverse, kelvin inverse, that is the r value multiply by temperature 373 kelvin.
02:51
Now from here, moles of oxygen are equal to 0 .044 mole.
03:00
Now we have to calculate the limiting reactant to produce the maximum no2...
