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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?

Grade:Upto college level

2 Answers

vikas askiitian expert
509 Points
10 years ago

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%

 

kiran kumar pidikiti
15 Points
10 years ago

                              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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