Question

How many degree for 2-methyl butene and 2-pentene are -30 kcal/mol and -28 kcal/mol respectively. The heats of combustion of 2-methyl butane and pentane are -784 kcal/mol and -782 kcal/mol respectively. All the value are given under standard conditions. Taking into account that combustion of both alkanes gives the same products, what is $\Delta H$ (in kcal/mol) for the following reaction under the same condition?

• A. 0
• B. -4
• C. -2
• D. 2

Answer 1

The correct option is D 2

$\Delta H_{reaction}=(\Delta H{)}_{products}-(\Delta H{)}_{products}$
$⟹\Delta H_{reaction}=(-782)-(-784)$
$⟹\Delta H_{reaction}=2$ $Kcal/mol$