The bond enthalpies of C-C, C=C, H-H and C-H | vedclass

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

Question

(C) The hydrogenation reaction of ethylene is: $CH_2=CH_2 + H_2 \rightarrow CH_3-CH_3$. The enthalpy change of the reaction is calculated as: $\Delta

Vedclass · Accepted Answer

$-180 \ kJ \ mol^{-1}$

Vedclass · Answer

(C) The hydrogenation reaction of ethylene is: $CH_2=CH_2 + H_2 ightarrow CH_3-CH_3$. The enthalpy change of the reaction is calculated as: $\Delta H = \sum 	ext{Bond Enthalpies of Reactants} - \sum 	ext{Bond Enthalpies of Products}$. Reactants: $1 	imes (C=C) + 4 	imes (C-H) + 1 	imes (H-H) = 600 + 4(410) + 400 = 600 + 1640 + 400 = 2640 \ kJ \ mol^{-1}$. Products: $1 	imes (C-C) + 6 	imes (C-H) = 360 + 6(410) = 360 + 2460 = 2820 \ kJ \ mol^{-1}$. $\Delta H = 2640 - 2820 = -180 \ kJ \ mol^{-1}$.

The bond enthalpies of $C-C, C=C, H-H$ and $C-H$ bonds are $360, 600, 400$ and $410 \ kJ \ mol^{-1}$ respectively. What is the heat of hydrogenation of ethylene?

Similar Questions

The heat of neutralization of four acids $A$,$B$,$C$,and $D$ are $-13.0$,$-12.6$,$-9.2$,and $-11.7 \ KCal/eq$ respectively when neutralized by $NaOH$. The order of acidic strength of the four acids will be:

Which of the following statements is correct about the heat of combustion?

Compounds with high heat of formation are less stable because

Explain the change in enthalpy related to a chemical reaction.

$PbO$ exists in two crystalline forms: yellow and red. The standard enthalpies of formation for these two forms are $-217.3 \ kJ/mol$ and $-219.0 \ kJ/mol$,respectively. Calculate the enthalpy of transition for the process: $PbO \text{ (yellow)} \rightarrow PbO \text{ (red)}$ in $kJ/mol$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module