Question

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)

Answer 1

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 

Answer 2

Kp = 1.483 . 10³

Further 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

$\large {\boxed {\bold {Kc ~ = ~ \frac {[C] ^ m [D] ^ n} {[A] ^ p [B] ^ q}}}}$

While the equilibrium constant is based on partial pressure

$\large {\boxed {\bold {Kp ~ = ~ \frac {[pC] ^ m [pD] ^ n} {[pA] ^ p [pB] ^ q}}}}$

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

$\large {\boxed {\bold {Kp ~ = ~ Kc. (R.T) ^ {\Delta n}}}}$

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)

$\rm COF_2(g)--->\dfrac{1}{2}CO_2(g)+\dfrac{1}{2}CF_4(g) (reaction\:2)$

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

$\rm Kp=\sqrt{2.2.10^6}\\\\Kp=1.483.10^3$

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