The reaction below has an equilibrium constant kp=2.2×106 at 298 k. 2cof2(g)⇌co2(g)+cf4(g) calculate kp for the reaction below. cof2(g)⇌12co2(g)+12cf4(g)

I believe the correct 2nd reaction is:

cof2(g)⇌1/2 co2(g)+1/2 cf4(g)

where we can see that it is exactly one-half of the original

Therefore the new Kp is:

new Kp = (old Kp)^(1/2)

new Kp = (2.2 x 10^6)^(1/2)

new Kp = 1,483.24 

The equilibrium constant is the ratio of concentration or pressure between the results of the reaction / product and the reactant with each reaction coefficient raised

The equilibrium constant is based on the concentration (Kc) in a reaction

pA + qB -----> mC + nD

While the equilibrium constant is based on partial pressure

The value of Kp and Kc can be linked to the formula '

R = gas constant = 0.0821 L.atm / mol.K

=n = number of product coefficients-number of reactant coefficients

There are rules in determining the Kp value for the related reaction

2COF₂ (g) ⇌CO₂(g) + CF₄ (g) Kp = 2.2 × 10⁶ at 298 K(reaction 1)

For the reaction 2, the reaction coefficient is divided by 2, so the new Kp value :

an equilibrium constant link

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Calculate the value of the equilibrium constant, Kc

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Concentration of hi at equilibrium

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let that person have a turn to speak : )