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for the reaction XCO3(s)======>XO(s) + CO2(g) Kp=1.642 at 727 degree celsius. If 4 moles XCO3(s) was put into a 50 litre container and heated to 727 degree celsius. what mole percent of the XCO3 remains unreacted at equilibrium?
XCO3 ==> XO + CO2 4 0 0 (initial) 4-x x x partial pressure of CO2 = p then Kp = p (at equilibrium ) pv = nRT p = nRT/V = xRT/V so Kp = XRT/V after plugging all the values we get X = 1 moles of XCO3 remaining are 3 ...mol % remain unreacted is 75%
XCO3 ==> XO + CO2
4 0 0 (initial)
4-x x x
partial pressure of CO2 = p then
Kp = p (at equilibrium )
pv = nRT
p = nRT/V = xRT/V so
Kp = XRT/V
after plugging all the values we get
X = 1
moles of XCO3 remaining are 3 ...mol % remain unreacted is 75%
XCO3(s)======>XO(s) + CO2(g) GIVEN INITIALLY 4 MOLES 0 0 MOLES AT EQ 4-a a a moles kp=1.642 => kp=Pco2 = 1.642 therefore acc toPV=nRT n = 1 mole therefore unreacted xco3= 4-a =4-1 = 3 mole so mole percent of unreacted reactant = 3 x100 /4 = 75%
XCO3(s)======>XO(s) + CO2(g)
GIVEN INITIALLY 4 MOLES 0 0 MOLES
AT EQ 4-a a a moles
kp=1.642 => kp=Pco2 = 1.642
therefore acc toPV=nRT n = 1 mole
therefore unreacted xco3= 4-a =4-1 = 3 mole
so mole percent of unreacted reactant = 3 x100 /4 = 75%
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