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Sagot :
The equilibrium partial pressure of [tex]CO_{2}[/tex] is 0.709 atm and CO is 0.262 atm.
What is equilibrium pressure?
The pressure that a vapour exerts when it is in thermodynamic equilibrium with the condensed phase (solid or liquid) in a closed system at a specific temperature is known as the vapour pressure or equilibrium vapour pressure. An indicator of a liquid's evaporation rate is the equilibrium vapour pressure.
How does partial pressure affect equilibrium?
The position of equilibrium is adjusted to reduce its partial pressure when the partial pressure of any of the gaseous reactants or products increases. This is often accomplished by favouring the reaction that results in a reduction in the number of gaseous components by moles.
19.9 = [tex]\frac{x^{3}}{(0.971-x)^{3}}[/tex]
Taking cubic root
2.71 = [tex]\frac{x}{0.971-x}[/tex]
2.63 - 2.71x = x
x = 0.709 atm = partial pressure [tex]CO_{2}[/tex]
Partial pressure CO = 0.971 - 0.709
= 0.262 atm
To know more about equilibrium vapour pressure, visit: https://brainly.com/question/13351348
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The complete question is:
“What are the equilibrium partial pressures of CO and [tex]CO_{2}[/tex] if CO is the only gas present initially, at a partial pressure of 0.938 atm?
At 1000 K, Kp= 19.9 for the reaction [tex]Fe_{2}O_{3}[/tex] + 3CO → 2Fe + 3 [tex]CO_{2}[/tex]“
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