Question

Butane exists as two isomers, $n$-butane and iso-butane

at $25^{\circ} C$
What is the pressure of iso-butane in a container of two isomer at equilibrium with a total pressure of $1.22\, atm ?$

• A. 0.61 atm
• B. 0.035 atm
• C. 1.00 atm
• D. 0.87 atm

Answer 1

Answer: (D)

Let pressure due to $n$ -butyne $=x\,atm$

Then pressure due to iso-butane $=(1.22-x)$

$K _{ p }=\frac{ p _{\text {iso-butane }}}{ p _{ n -\text { butyne}}}$

$\frac{1.22-x}{x}=2.5$

$1.22-x=2.5\, x$

$3.5 x=1.22$

$x=0.35\, atm$

Thus, $p _{\text {butane }}=0.35\, atm$

$p _{\text {iso-butane }}=1.22-0.35=0.87\, atm$