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 

Kp = 1.483 . 10³Further explanation

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

1. If the reaction equation is reversed, the value of Kp is also reversed 2. if the reaction coefficient is divided by a factor n, then the new Kp value is the square root of n from the old Kp 3. if the reaction coefficient is multiplied by the factor n, then the new Kp value is the power of n from the old Kp

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 :

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