The equilibrium constant for the reaction,

H2(g) + I2(g) == 2 HI(g) is 54.9 at 699.0 K (Kelvin).
What is the equilibrium constant for
4 HI(g) == 2 H2(g) + 2 I2(g) under the same conditions?

Note: the == indicates the equilibrium double arrow

Since the second equation is doubled I multiplied 54.9 by 2 and got 109.8. Is this correct??

No.

for the second equation, try this

kc=(1/54.9)^2

if you cannot figure out why, state so.

I do not understand why do you 1/54.9??

You take 1/2 of an equation, new k = sqrt old k.

You double an equation, new k = old k^2
You revere an equation, new k = 1/old k.

To determine the equilibrium constant for the second reaction, you need to use the relationship between the equilibrium constants and the stoichiometric coefficients of the balanced equation.

In this case, the reaction equation you provided is:

4 HI(g) ⇌ 2 H2(g) + 2 I2(g)

The balanced equation shows that for every 4 moles of HI consumed, 2 moles of H2 and 2 moles of I2 are produced.

The equilibrium constant expression is given by:

K = [H2]^2 [I2]^2 / [HI]^4

Now, using the equilibrium constant from the first reaction (K = 54.9), we have:

54.9 = [HI]^2 / ([H2][I2])

Since the stoichiometric coefficients for the second reaction are twice those in the first reaction, we can rewrite the concentration of H2 and I2 as ([H2] / 2) and ([I2] / 2), respectively.

Substituting these values into the equilibrium constant expression, we get:

K' = [H2]^2 [I2]^2 / [HI]^4
= ([H2] / 2)^2 ([I2] / 2)^2 / ([HI] / 2)^4
= [H2]^2 [I2]^2 / [HI]^4
= (2^2) (2^2) (2^4) [H2]^2 [I2]^2 / [HI]^2 [HI]^2
= 2^8 [H2]^2 [I2]^2 / ([HI]^4)
= 256 [H2]^2 [I2]^2 / ([HI]^4)

Now, we substitute the expression for K from the first reaction:

K' = 256 * 54.9
= 14079.4

Therefore, the equilibrium constant for the reaction 4 HI(g) ⇌ 2 H2(g) + 2 I2(g) under the same conditions is approximately 14079.4.