the equilibrium constant Kp for thus reaction is 4. at 1660°C. Initially 0.8 mole H2 and 0.8 mole of CO2 are injected into a 5 L flask. what will be the equilibrium constant for CO2??
Answers
Moles of both sides are equal So according to relation between kp and kc ...
When moles of product are equal to reactant moles
Kp = kc
Hence
Kc = kp = 4
Kc =4
H2. +co2. -------> h20. + co
Initial 0.8. 0.8. 0. 0
Equil:(0.8- x). (0.8 -x) x. X
Kc = product ÷ reactant
Kc = (x) (x) ÷( 0.8-x ) (0.8- x )
4 = X2 ÷ (0.64 _0.16x +X2)
4 ( 0.64 _0.16x +X2) = X2
2.56 - 0.64x + 4X2 = X2
2.56 - 0.64x + 4X2 - X2 = 0
2.56 - 0.64x + 3X2 =0
Then use quadratic farmula ....
Hope it helps you
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Answer:
Explanation:
Kp = kc
Hence
Kc = kp = 4
Kc =4
H2. +co2. -------> h20. + co
Initial 0.8. 0.8. 0. 0
Equil:(0.8- x). (0.8 -x) x. X
Kc = product ÷ reactant
Kc = (x) (x) ÷( 0.8-x ) (0.8- x )
4 = X2 ÷ (0.64 _0.16x +X2)
4 ( 0.64 _0.16x +X2) = X2
2.56 - 0.64x + 4X2 = X2
2.56 - 0.64x + 4X2 - X2 = 0
2.56 - 0.64x + 3X2 =0
Then use quadratic farmula ....
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