Question

A gaseous mixture of 10.0 volumes of CO₂, 1.00 volume of unreacted O₂, and 50.0 volumes of unreacted N₂ leaves an engineat 4.0 atm and 800. K. Assuming that the mixture reaches equilibrium, what are (b) the concentration (in picograms per liter, pg/L) of CO in this exhaust gas?2CO₂(g) ⇄ 2CO(g) + O₂(g) Kp = 1.4×10⁻²⁸ at 800K(The actual concentration of CO in exhaust gas is much higher because the gases do not reach equilibrium in the short transit time through the engine and exhaust system.)

Answer 1

The masses of CO and CO2​ are 90.55g and 100−90.55=9.45 g respectively.

Total mass.

Let the mixture has 100g as total mass.

The number of moles of CO is 2890.55​=3.234.

The number of moles of CO2​ is 449.45​=0.215.

The mole fraction of CO is 3.234+0.2153.234​=0.938.

The mole fraction of CO2​ is 1−0.938=0.062.

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

It is 0.938×1=0.938 atm.

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

The expression for the equilibrium constant is:

Kp​=PCO2​​PCO2​​=0.062(0.938)2​=14.19

Δng​=2−1=1

Kc​=Kp​(RT)−Δn=14.19×(0.0821×1127)−1=0.153.

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