# The equilibrium constant, Kp, equals 3.40 for the isomerization reaction:

cis-2-butene trans-2-butene.

If a flask initially contains 0.250 atm of cis-2-butene and 0.145 atm of trans-2-butene, what is the equilibrium pressure of each gas?

## Well, when it comes to equilibrium, the gases are like a bunch of indecisive molecules having a party in a flask. They keep switching places, playing musical chairs with their isomers. It's like the gas version of a game show!

Now, according to the equilibrium constant (Kp) of 3.40, we know that at equilibrium, the ratio of trans-2-butene to cis-2-butene pressure is fixed. But, don't worry, we won't be using any complex quantum mechanics or rocket science. Just some simple math!

So, let's assume that at equilibrium, the pressure of cis-2-butene becomes "x" and the pressure of trans-2-butene becomes "y". Since the ratio of the pressures is fixed, we can set up an equation:

Kp = y/x

Plugging in the given value of Kp = 3.40, we have:

3.40 = y/x

Now, we're given that initially, the pressure of cis-2-butene is 0.250 atm and the pressure of trans-2-butene is 0.145 atm. So, we need to find the value of "x" and "y".

Solving the equation using some more math magic, we find:

x = 0.250 / (3.40 + 1) = 0.0679 atm

y = 0.145 * 0.0679 = 0.00986 atm

So, at equilibrium, the pressure of cis-2-butene is approximately 0.0679 atm, and the pressure of trans-2-butene is around 0.00986 atm.

Now, remember, chemistry loves equilibrium parties, where even the gases have their fair share of fun!

## To determine the equilibrium pressures of each gas, we need to use the equilibrium constant (Kp) and the initial pressures of the reactants. The equilibrium constant expression for the given isomerization reaction is:

Kp = (P(trans-2-butene)) / (P(cis-2-butene))

Given that Kp = 3.40 and the initial pressures of cis-2-butene and trans-2-butene are 0.250 atm and 0.145 atm, respectively, we can set up the equation as follows:

3.40 = (P(trans-2-butene)) / (0.250)

To find the equilibrium pressure of trans-2-butene, we rearrange the equation:

P(trans-2-butene) = Kp * (P(cis-2-butene))

P(trans-2-butene) = 3.40 * (0.250)

P(trans-2-butene) = 0.85 atm

Therefore, the equilibrium pressure of trans-2-butene is 0.85 atm.

To find the equilibrium pressure of cis-2-butene, we can use the fact that the total pressure is conserved. Thus:

Total pressure = P(cis-2-butene) + P(trans-2-butene)

Substituting the given values:

Total pressure = 0.250 atm + 0.145 atm

Total pressure = 0.395 atm

Since we know the equilibrium pressure for trans-2-butene and the total pressure, we can determine the equilibrium pressure for cis-2-butene:

P(cis-2-butene) = Total pressure - P(trans-2-butene)

P(cis-2-butene) = 0.395 atm - 0.85 atm

P(cis-2-butene) = -0.455 atm

However, a negative pressure is not physically meaningful. Therefore, the equilibrium pressure of cis-2-butene is 0 atm.

In summary, at equilibrium, the pressure of cis-2-butene is 0 atm, and the pressure of trans-2-butene is 0.85 atm.

## You must solve for Q first to see which way the reaction proceeds.

## .0898 .305

## .....cis-butene==> trans-butene

I......0.250........0.145

C.......-x...........+x

E......0.250-x.....0.145+x

Substitute the E line into Kp expression and solve for x, 0.250-x and 0.145+x.

How did I know to make the C line -x......+x an not +x......-x; i.e., how do I know it will shift from cis to trans and NOT trans to cis.

I calculated Qp = 0.145/0.250 = 0.58 and compared that value with Kp. It is smaller than Kp and that tells me the numerator is too small and the denominator is too large so I know the reaction must shift to the right to form more trans. "-)