The heats of hydrogenation for 3--methylbuten | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let the reaction be: $3-$-methylbutene $\rightarrow$ $2-$-pentene.We can use Hess's Law by constructing a cycle:$1.$ $3-$-methylbutene $+ H_2 \rig

Vedclass · Accepted Answer

$-4$

Vedclass · Answer

(B) Let the reaction be: $3-$-methylbutene $ightarrow$ $2-$-pentene.We can use Hess's Law by constructing a cycle:$1.$ $3-$-methylbutene $+ H_2 ightarrow 2-$-methylbutane,$\Delta H_1 = -30\, kcal/mol$$2.$ $2-$-methylbutane $ightarrow$ pentane,$\Delta H_2 = \Delta H_{comb}(2-$-methylbutane$) - \Delta H_{comb}(	ext{pentane}) = -784 - (-782) = -2\, kcal/mol$$3.$ Pentane $ightarrow 2-$-pentene $+ H_2$,$\Delta H_3 = -(\Delta H_{hydrog}(2-$-pentene$)) = -(-28) = +28\, kcal/mol$Adding these steps:$3-$-methylbutene $ightarrow 2-$-pentene$\Delta H = \Delta H_1 + \Delta H_2 + \Delta H_3 = -30 - 2 + 28 = -4\, kcal/mol$.

Similar Questions

Which of the following has standard enthalpy of formation equal to zero?

$XeF_{2(g)} + H_{2(g)} \to 2HF_{(g)} + Xe_{(g)}$,$\Delta H^o = -430 \ kJ$
Bond energy:
$H-H = 435 \ kJ/mol$
$H-F = 565 \ kJ/mol$
Calculate the average bond energy of the $Xe-F$ bond in $kJ/mol$.

Write notes on diatomic molecules and polyatomic molecules with respect to bond dissociation enthalpy.

If the enthalpies of combustion of benzene $(l)$,carbon $(s)$,and hydrogen $(g)$ are $Q_1$,$Q_2$,and $Q_3$ respectively,what will be the enthalpy of formation of benzene?

Vedclass Test Series

Exam Paper Generator

Online Exam Module