Question

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??

Answer 1

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 ....


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Answer 2

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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