Question

Enthalpy of combustion of propane,butane and pentane are -2220, -2878 , -3537 kJ/mol respectively. Order of chlorific value ( per gram) of fuel will be.
1. propane > Butane > pentane.
2. Butane > Pentane > propane.
3. Propane < butane < pentane.
4. pentane < butane < propane​

Answer 1

Answer:

The correct answer is option 3.

Explanation:

Calorific value of a substance is defined as amount of heat generated on combustion of a fixed amount of a substance. Generally expressed in Joules per kilograms.

Enthalpy of combustion of propane = -2220 kJ/mol

Mass of 1 mol = 44 g

Calorific value of propane:

$\frac{-2220 kJ/}{ 44 g}=-50.45 kJ/g$

Enthalpy of combustion of butane = -2878 kJ/mol

Mass 1 mol = 58 g

Calorific value of butane:

$\frac{-2878 kJ/}{ 58 g}=-49.62 kJ/g$

Enthalpy of combustion of pentane = -3537 kJ/mol

Mass of 1 mol = 72 g

Calorific value of pentane :

$\frac{-3537 kJ/}{ 72 g}=-49.125 kJ/g$

Order of calorific value ( per gram) of fuel will be:

Propane < butane < pentane

Answer 2

Answer:

1 is the correct answer

Explanation:

calorific value of propane = - 50.45kj/g

butane. = - 49.62kj/g

pentane = - 49.12kj/g

here we wants to consider magnitude only..

hence propane > butane > pentane