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 $1.22atm?$

• 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 $=xatm$

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

$K_p=\frac{p_{\text{iso-butane }}}{p_{n-\text{ butyne}}}$

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

$1.22-x=2.5x$

$3.5x=1.22$

$x=0.35atm$

Thus, $p_{\text{butane }}=0.35atm$

$p_{\text{iso-butane }}=1.22-0.35=0.87atm$