From the following bond energies : H-H bond energy : 431.37 ~kJ ~mol^ -1 C=C bond energy : 606.10 ~kJ ~mol^ -1 C - C bond energy : 336.49 ~kJ ~mol^ -1 C-H bond energy : 410.50 ~kJ ~mol^ -1 Enthalpy for the reaction

Correct answer: C. -120.0 ~kJ ~mol^ -1 Key Idea Enthalpy of reaction = ∑ BE_ Reactants -∑ BE_ Products For reaction, aligned & Δ H_ r =[4 × BE_ (C-H) +1 × BE_ (C=C) +1 × BE_ (H-H) ] & -[6 × BE_ (C-H) +1 × BE_ (C-C) ] & =(4 ×410.50+1 ×606.10+1 ×431.37) & -[(6 ×410.50)+(1 ×336.49)] & =-120.0 ~kJ ~mol^ -1 aligned

NEET 2009

Exam: NEET
Year: 2009
Subject: Chemistry
Chapter: Thermodynamics (C)

NEET 2009 · Chemistry · Thermodynamics (C)

From the following bond energies : H-H bond energy : 431.37 ~kJ...

From the following bond energies :

\mathrm{H}-\mathrm{H}

bond energy :

431.37 \mathrm{~kJ} \mathrm{~mol}^{-1}

\mathrm{C}=\mathrm{C}

bond energy :

606.10 \mathrm{~kJ} \mathrm{~mol}^{-1}

\mathrm{C}

\mathrm{C}

bond energy :

336.49 \mathrm{~kJ} \mathrm{~mol}^{-1}

\mathrm{C}-\mathrm{H}

bond energy :

410.50 \mathrm{~kJ} \mathrm{~mol}^{-1}

Enthalpy for the reaction, From the following bond energies : H-H bond energy : 431.37 ~kJ ~mol^ -1 C=C bond energy : 606.10 ~kJ ~mol^ -1 C - C

will be

Options

A $1523.6 \mathrm{~kJ} \mathrm{~mol}^{-1}$
B $-243.6 \mathrm{~kJ} \mathrm{~mol}^{-1}$
C $-120.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$
D $553.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Correct answer

C. $-120.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Step-by-step solution

Key Idea Enthalpy of reaction =

\sum \mathrm{BE}_{\text{Reactants }}-\sum \mathrm{BE}_{\text{Products }}

For reaction,

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\begin{aligned}& \Delta \mathrm{H}_{\mathrm{r}}=\left[4 \times \mathrm{BE}_{(\mathrm{C}-\mathrm{H})}+1 \times \mathrm{BE}_{(\mathrm{C}=\mathrm{C})}+1 \times \mathrm{BE}_{(\mathrm{H}-\mathrm{H})}\right]\\ & -\left[6 \times \mathrm{BE}_{(\mathrm{C}-\mathrm{H})}+1 \times \mathrm{BE}_{(\mathrm{C}-\mathrm{C})}\right]\\ & =(4 \times410.50+1 \times606.10+1 \times431.37) \\ & \quad-[(6 \times410.50)+(1 \times336.49)] \\ & =-120.0 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned}

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