Some Basic Concepts Of Chemistry Question 132
Question: 23.2 g of an organic compound having molecular formula $ C _{n}{H _{2n+2}} $ is burnt in excess of $ O _2(g) $ initially taken in a 44.82 L steel vessel. Before reaction the gaseous mixture w at 273 K with pressure of 2 aim. After complete combustion and loss of considerable amount of heat the mixture of product and excess of had a temperature of 546 K and 4.6 atm pressure. The formula of compound is
Options:
A) $ C _5H _{12} $
B) $ C _6H _{10} $
C) $ C _3H _8 $
D) $ C _4H _{10} $
Show Answer
Answer:
Correct Answer: D
Solution:
[d] Idea This problem can be solved by using concept of ideal gas equation and chemical equation involved in combustion reaction. Students are advised to follow steps. Write chemical equation involved in combustion of hydrocarbon. Assume initial pressure = p Calculate increased pressure Then, calculate number of mole using ideal gas equation. Equation of combustion $ C _{n}H _2n _{+2}+( \frac{3n+1}{2} )O _2\xrightarrow{{}}nCO _2+(n+1)H _2O $ Initial pressure of $ C _{n}H _2n _{+2} $ is p (assumed) Increase in pressure $ =p[ (2n+1)-1-( \frac{3n+1}{2} ) ]=( \frac{n-1}{2} )p $
$ \because $ Mass of organic compound $ =14\times n+2 $
546 K and 4.6 atm or 273 K 2.3 atm
Increase
$ \Rightarrow $ $ 2.3-2=0.3 $ atm
$ p=\frac{nRT}{V}=\frac{23.2}{M}\times ( \frac{0.0821\times 273}{44.82} ) $
$ =\frac{23.2}{M}\times \frac{0.5}{44.82} $
$ =\frac{116}{(14n+2)}\times \frac{(n-1)}{2} $ also $ =\frac{(n-1)}{2}\times \frac{11.6}{14n+2}=0.3 $ On solving, $ n=4 $ So, compound will be $ C _4{H _{4\times 2+2}}\Rightarrow C _4H _{10} $
