Question

For the gaseous reaction involving complete combustion of isobutane (assuming all products and reactants are in gaseous state), the relation between\[\Delta H\]and \[\Delta E\] will be:
A. \[\Delta H>\Delta E\]
B. \[\Delta H=\Delta E\]
C. \[\Delta H<\Delta E\]
D. \[\Delta H=0and\Delta E=\infty \]

Answer 1

Hint: Butane is an organic compound which is saturated hydrocarbon. It is widely used as a gasoline mixture and as a feedstock for ethylene. On oxidation it tends to form carbon dioxide and water on reaction with chlorine it forms butyl chloride and hydrochloric acid. It is also used as a fuel gas.

Complete answer:
To approach this question we need to write the reaction of isobutane with the oxygen that means we need to show the oxidation of isobutane. So as I said that on oxidation it forms carbon dioxide and water molecules. So the reaction for them is the following: \[{{C}_{4}}{{H}_{10}}(g)+\dfrac{3}{2}{{O}_{2}}\Delta H=\Delta E+\Delta nRT(g)\to 4C{{O}_{2}}(g)+5{{H}_{2}}O(g)\]

The relation between \[\Delta H\] and \[\Delta E\] is the following:
\[\Delta H=\Delta E+\Delta nRT\]
- Where the \[\Delta n\] is the number of gaseous moles in the product subtracted by the number of gaseous moles in the reactant.
- So \[\Delta n\] will be equal to = (4+5)-(1+\[\dfrac{13}{2}\])
- So \[\Delta n\] = \[\dfrac{3}{2}\](+ve)
Since we see that\[\Delta n\] is positive so the \[\Delta H\] will be greater than the value for \[\Delta E\] because the addition of \[\Delta E\] and \[\Delta nRT\]will be equal to \[\Delta H\].

Hence the correct answer is option A.

Note: Butane is considered as a petroleum derived gaseous liquid which is widely used for camping, cooking and in cigarette lighters. It is also used for the production of MTBE that is methyl tertiary butyl ether. It is also used in the catalytic dehydrogenation of butane to butenes. It is used in the domestic cylinders which are painted red.