Answer :
Answer:
Pco = 1.51 atm
Pco[tex]_{2}[/tex] = 1.5 atm
Explanation:
C (s) + CO2 (g) ⇔ 2 CO (g)
T = 700°C, P (total) = 3.01 atm, KP = 1.52
Mathematically,
KP = [tex]\frac{(Pco)^{2} }{Pco_{2} }[/tex]
According to Dalton' Law which states that at equilibrium, the sum of the partial pressures of each component is equal to the total pressure
Mathematically,
P (total) = Pco[tex]_{2}[/tex] + Pco = 3.01 atm
Pco[tex]_{2}[/tex] = P (total) - Pco
Pco[tex]_{2}[/tex] = 3.01 - Pco
Substitute Pco[tex]_{2}[/tex] into the KP equation, we have
KP = [tex]\frac{(Pco)^{2} }{3.01 - Pco }[/tex] ⇒ 1.52 = [tex]\frac{(Pco)^{2} }{3.01 - Pco }[/tex]
Let y = Pco
1.52 = [tex]\frac{(y)^{2} }{3.01 - y }[/tex] = [tex]y^{2}[/tex] = 1.52 (3.01 - y)
[tex]y^{2}[/tex] = 4.5752 - 1.52y ⇒ [tex]y^{2}[/tex] + 1.52y - 4.5752 = 0
Using the quadratic formula y = [tex]\frac{-b + \sqrt{b^{2} -4ac } }{2a}[/tex] and y = [tex]\frac{-b - \sqrt{b^{2} -4ac } }{2a}[/tex]
y = 1.51 or −3.03
We choose the positive root because Pco cannot be negative (Remember Pco = y)
Hence, Pco = 1.51 atm
Pco[tex]_{2}[/tex] = 3.01 - Pco = 3.01 - 1.51
Pco[tex]_{2}[/tex] = 1.5 atm