Consider the isomerization of butane with equilibrium constant is 2.5 .The system is originally at equilibrium with :
[butane]=1.0 M , [isobutane]=2.5 M
If 0.50 mol/L of butane is added to the original equilibrium mixture and the system shifts to a new equilibrium position, what is the equilibrium concentration of each gas?
Answer:
The equilibrium concentration of each gas:
[Butane] = 1.14 M
[isobutane] = 2.86 M
Explanation:
Butane ⇄ Isobutane
At equilibrium
1.0 M 2.5 M
After addition of 0.50 M of butane:
(1.0 + 0.50) M -
After equilibrium reestablishes:
(1.50-x)M (2.5+x)
The equilibrium expression will wriiten as:
![K_c=\frac{[Isobutane]}{[Butane]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BIsobutane%5D%7D%7B%5BButane%5D%7D)

x = 0.36 M
The equilibrium concentration of each gas:
[Butane]= (1.50-x) = 1.50 M - 0.36M = 1.14 M
[isobutane]= (2.5+x) = 2.50 M + 0.36 M = 2.86 M
The one with higher temperature is the one with NaOH as heatis given off during the neutralization reaction that occurs.
<h3>
What is volume?</h3>
Volume can be defined as the amount of space a substance or an objects occupies usually in a closed container.
Volume is measures in litres.
When water is added to dilute acid like HCl, they become more dilute.
When NaOH is added to HCl, a neutralization reaction occurs.
The student will determine the contents of the flasks by adding 10 ml of hcl to each flask. If the NaOH reacts with the Hcl, there will be an increase in temperature.
The increase in temperature is due to the heat of neutralization of the reaction between NaOH and HCl.
Learn more about volume at: brainly.com/question/1972490
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The answer is mean,mode and median