A gaseous mixture of 10.0 volumes of COโ‚‚, 1.00 volume of unreacted Oโ‚‚, and 50.0 volumes of unreacted Nโ‚‚ leaves an engineat 4.0 atm and 800. K. Assuming that the mixture reaches equilibrium, what are (a) the partial pressure and

Respuesta :

The masses of CO and CO2โ€‹ are 90.55g and 100โˆ’90.55=9.45 g respectively.

Total mass.

Let the mixture has 100g as total mass.

The number of moles of CO is 2890.55โ€‹=3.234.

The number of moles of CO2โ€‹ is 449.45โ€‹=0.215.

The mole fraction of CO is 3.234+0.2153.234โ€‹=0.938.

The mole fraction of CO2โ€‹ is 1โˆ’0.938=0.062.

The partial pressure of CO is the product of the mole fraction of CO and the total pressure.

It is 0.938ร—1=0.938 atm.

The partial pressure of carbon dioxide is 0.062ร—1=0.042 atm.

The expression for the equilibrium constant is:

Kpโ€‹=PCO2โ€‹โ€‹PCO2โ€‹โ€‹=0.062(0.938)2โ€‹=14.19

ฮ”ngโ€‹=2โˆ’1=1

Kcโ€‹=Kpโ€‹(RT)โˆ’ฮ”n=14.19ร—(0.0821ร—1127)โˆ’1=0.153.

To learn more about equilibrium constant visit the link

https://brainly.com/question/15118952

#SPJ4