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?

Question

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?

vikas askiitian expert · Answer

XCO 3 ==> XO + CO 2 4 0 0 (initial) 4-x x x partial pressure of CO 2 = p then K p = p (at equilibrium ) pv = nRT p = nRT/V = xRT/V so K p = XRT/V after plugging all the values we get X = 1 moles of XCO 3 remaining are 3 ...mol % remain unreacted is 75%

kiran kumar pidikiti · Answer

XCO3(s)======>XO(s) + CO2(g)

GIVEN INITIALLY 4 MOLES 0 0 MOLES

AT EQ 4-a a a moles

k_p=1.642 => k_p=P_co2= 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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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?

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?

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