Question

Following reaction is given $CH _{3} COCH _{3}( g ) \rightleftharpoons CH _{3}- CH _{3}+ CO ( g ),$ initial pressure of $CH _{3} COCH _{3}$ is $100 \,mm$ of $Hg$. When equilibrium is set up, the mole fraction of $CO _{( g )}$ is $1 / 3,$ hence $K _{ p }$ is-

• A. 10 mm
• B. 50 mm
• C. 25 mm
• D. 150 mm

Answer 1

Answer: (B)

Since, rest all parametess are constant (v, T), we can say that

$P_{C O_{(at \,eqm)}}=\frac{1}{3} P_{\text {Total }_{(at\, eqm)}}$

$P=\frac{1}{3}(100-P+P+P)$

$ \Rightarrow P=\frac{1}{3}(100+P)$

$3 p=100+p$

$ \Rightarrow 2 p=100$

$p=50\,mm$ of $Hg$

$K _{ p }=\frac{\left( pCH _{3}- CH _{3}\right)( pCO )}{\left( pCH _{3} COCH _{3}\right)}=\frac{50 \times 50}{(100-50)}=\frac{50 \times 50}{50}$

$=50 \,mm$